ch15 (1)

15.1 SOLUTIONS 1039

CHAPTER FIFTEEN

Solutions for Section 15.1

Exercises

1. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point and is a localmaximum; the point C is a saddle point.

2. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving

fx = 2x− 2y = 0,

fy = −2x+ 6y − 8 = 0.

We see from the first equation that x = y. Substituting this into the second equation shows that y = 2. The only criticalpoint is (2, 2).

We haveD = (fxx)(fyy)− (fxy)2 = (2)(6)− (−2)2 = 8.

Since D > 0 and fxx = 2 > 0, the function f has a local minimum at the point (2, 2).

3. The partial derivatives arefx = −6x− 4 + 2y and fy = 2x− 10y + 48.

Set fx = 0 and fy = 0 to find the critical point, thus

2y − 6x = 4 and 10y − 2x = 48.

Now, solve these equations simultaneously to obtain x = 1 and y = 5.Since fxx = −6, fyy = −10 and fxy = 2 for all (x, y), at (1, 5) the discriminant D = (−6)(−10)− (2)2 = 56 >

0 and fxx < 0. Thus f(x, y) has a local maximum value at (1, 5).


fx = 3x2 − 6x = 0

fy = 2y + 10 = 0

shows that x = 0 or x = 2 and y = −5. There are two critical points: (0,−5) and (2,−5).We have

D = (fxx)(fyy)− (fxy)2 = (6x− 6)(2)− (0)2 = 12x− 12.

When x = 0, we have D = −12 < 0, so f has a saddle point at (0,−5). When x = 2, we have D = 12 > 0 andfxx = 6 > 0, so f has a local minimum at (2,−5).


fx = 3x2 − 3 = 0

fy = 3y2 − 3 = 0

shows that x = ±1 and y = ±1. There are four critical points: (1, 1), (−1, 1), (1,−1) and (−1,−1).We have

D = (fxx)(fyy)− (fxy)2 = (6x)(6y)− (0)2 = 36xy.

At the points (1,−1) and (−1, 1), we have D = −36 < 0, so f has a saddle point at (1,−1) and (−1, 1). At (1, 1),we have D = 36 > 0 and fxx = 6 > 0, so f has a local minimum at (1, 1). At (−1,−1), we have D = 36 > 0 andfxx = −6 < 0, so f has a local maximum at (−1,−1).

1040 Chapter Fifteen /SOLUTIONS


fx = 3x2 − 6x = 0

fy = 3y2 − 3 = 0

shows that x = 0 or x = 2 and y = −1 or y = 1. There are four critical points: (0,−1), (0, 1), (2,−1), and (2, 1).We have

D = (fxx)(fyy)− (fxy)2 = (6x− 6)(6y)− (0)2 = (6x− 6)(6y).

At the point (0,−1), we have D > 0 and fxx < 0, so f has a local maximum.At the point (0, 1), we have D < 0, so f has a saddle point.At the point (2,−1), we have D < 0, so f has a saddle point.At the point (2, 1), we have D > 0 and fxx > 0, so f has a local minimum.


fx = 3x2 − 3 = 0

fy = 3y2 − 12y = 0

shows that x = −1 or x = 1 and y = 0 or y = 4. There are four critical points: (−1, 0), (1, 0), (−1, 4), and (1, 4).We have

D = (fxx)(fyy)− (fxy)2 = (6x)(6y − 12)− (0)2 = (6x)(6y − 12).

At critical point (−1, 0), we have D > 0 and fxx < 0, so f has a local maximum at (−1, 0).At critical point (1, 0), we have D < 0, so f has a saddle point at (1, 0).At critical point (−1, 4), we have D < 0, so f has a saddle point at (−1, 4).At critical point (1, 4), we have D > 0 and fxx > 0, so f has a local minimum at (1, 4).

8. Find the critical point(s) by setting

fx = (xy + 1) + (x+ y) · y = y2 + 2xy + 1 = 0,

fy = (xy + 1) + (x+ y) · x = x2 + 2xy + 1 = 0,

then we get x2 = y2, and so x = y or x = −y.If x = y, then x2 +2x2 +1 = 0, that is, 3x2 = −1, and there is no real solution. If x = −y, then x2−2x2 +1 = 0,

which gives x2 = 1. Solving it we get x = 1 or x = −1, then y = −1 or y = 1, respectively. Hence, (1,−1) and (−1, 1)are critical points.Since

fxx(x, y) = 2y,

fxy(x, y) = 2y + 2x and

fyy(x, y) = 2x,

the discriminant is

D(x, y) = fxxfyy − f2xy

= 2y · 2x− (2y + 2x)2

= −4(x2 + xy + y2).

thus

D(1,−1) = −4(12 + 1 · (−1) + (−1)2) = −4 < 0,

D(−1, 1) = −4((−1)2 + (−1) · 1 + 12) = −4 < 0.

Therefore (1,−1) and (−1, 1) are saddle points.

9. At a critical point, fx = 0, fy = 0.

fx = 8y − (x+ y)3 = 0, we know 8y = (x+ y)3.

fy = 8x− (x+ y)3 = 0, we know 8x = (x+ y)3.

Therefore we must have x = y. Since (x+y)3 = (2y)3 = 8y3, this tells us that 8y−8y3 = 0. Solving gives y = 0,±1.Thus the critical points are (0, 0), (1, 1), (−1,−1).

15.1 SOLUTIONS 1041

fyy = fxx = −3(x+ y)2, and fxy = 8− 3(x+ y)2.The discriminant is


= 9(x+ y)4 −(64− 48(x+ y)2 + 9(x+ y)4

)

= −64 + 48(x+ y)2.

D(0, 0) = −64 < 0, so (0, 0) is a saddle point.D(1, 1) = −64 + 192 > 0 and fxx(1, 1) = −12 < 0, so (1, 1) is a local maximum.D(−1,−1) = −64 + 192 > 0 and fxx(−1,−1) = −12 < 0, so (−1,−1) is a local maximum.


fx = 6− 2x+ y = 0,

fy = x− 2y = 0.

We see from the second equation that x = 2y. Substituting this into the first equation shows that y = 2. The only criticalpoint is (4, 2).

We haveD = (fxx)(fyy)− (fxy)2 = (−2)(−2)− 12 = 3.

Since D > 0 and fxx = −2 < 0, the function f has a local maximum at (4, 2).


fx = e2x2+y2

(4x) = 0,

fy = e2x2+y2

(2y) = 0,

shows that the only critical point is (0, 0).We have

D = (fxx)(fyy)− (fxy)2 = e2x2+y2

(4 + (4x)2) · e2x2+y2

(2 + (2y)2)− (e2x2+y2

(4x · 2y))2.

At (0, 0), we have D = 4 · 2− 02 > 0 and fxx = 4 > 0, so the function has a local minimum at the point (0, 0).

12. At the origin f(0, 0) = 0. Since x6 ≥ 0 and y6 ≥ 0, the point (0, 0) is a local (and global) minimum. The secondderivative test does not tell you anything since D = 0.

13. At the origin g(0, 0) = 0. Since y3 ≥ 0 for y > 0 and y3 < 0 for y < 0, the function g takes on both positive and negativevalues near the origin, which must therefore be a saddle point. The second derivative test does not tell you anything sinceD = 0.

14. At the origin h(0, 0) = 1. Since cosx and cos y are never above 1, the origin must be a local (and global) maximum. Thesecond derivative test

D = hxxhyy − (hxy)2 =((− cosx cos y)(− cosx cos y)− (sinx sin y)2

) ∣∣x=0,y=0

=(cos2 x cos2 y − sin2 x sin2 y

) ∣∣x=0,y=0

= 1 > 0

and hxx(0, 0) < 0, so (0, 0) is a local maximum.

15. At the origin, the second derivative test gives

D = kxxkyy − (kxy)2 =((− sinx sin y)(− sinx sin y)− (cosx cos y)2

)∣∣x=0,y=0

= sin2 0 sin2 0− cos2 0 cos2 0

= −1 < 0.

Thus k(0, 0) is a saddle point.

Problems

16. (a) P is a local maximum.(b) Q is a saddle point.(c) R is a local minimum.(d) S is none of these.


17. Figure 15.1 shows the gradient vectors around P and Q pointing perpendicular to the contours and in the direction ofincreasing values of the function.

y

x

0

00

0

1234

5

6

11

23

45

6

6

−1

−1

−2

−3

−4

−5

−6

−1

−2

−3

−4

−5

−6−6QR

P

S

�:zRW�

�I

Yi�)�

II

��

RR

i1�-R

Figure 15.1

18. Figure 15.2 shows the direction of ∇f at the points where ‖∇f‖ is largest, since at those points the contours are closesttogether.

y

x

0

00

0

1234

5

6

112

345

6

−1

−1

−2

−3

−4

−5

−6

−1

−2

−3

−4

−5

−6−6

QR

P

S

Figure 15.2

19. First, we identify the critical points. The partials are fx(x, y) = 3x2 and fy(x, y) = −2ye−y2

. These will vanishsimultaneously when x = 0 and y = 0, so our only critical point is (0, 0). The discriminant is

D = fxx(x, y)fyy(x, y)− f2xy(x, y) = (6x)(4y2e−y

2 − 2e−y2

)− 0 = 12xe−y2

(2y2 − 1).

Unfortunately, the discriminant is zero at the origin so the second derivative test can tell us nothing about our critical point.We can, however, see that we are at a saddle point by looking at the behavior of f(x, y) along the line y = 0. Here we havef(x, 0) = x3 + 1, so for positive x, we have f(x, 0) > 1 = f(0, 0) and for negative x, we have f(x, 0) < 1 = f(0, 0).So f(x, y) has neither a maximum nor a minimum at (0, 0).

15.1 SOLUTIONS 1043

20. To find critical points, set partial derivatives equal to zero:

Ex = sinx = 0 when x = 0, ±π, ±2π, · · ·

Ey = y = 0 when y = 0.

The critical points are· · · (−2π, 0), (−π, 0), (0, 0), (π, 0), (2π, 0), (3π, 0) · · ·

To classify, calculate D = ExxEyy − (Exy)2 = cosx.At the points (0, 0), (±2π, 0), (±4π, 0), (±6π, 0), · · ·

D = (1) > 0 and Exx > 0 (SinceExx(0, 2kπ) = cos(2kπ) = 1).

Therefore (0, 0), (±2π, 0), (±4π, 0), (±6π, 0), · · · are local minima.At the points (±π, 0), (±3π, 0), (±5π, 0), (±7π, 0), · · ·, we have cos(2k + 1)π = −1, so

D = (−1) < 0.

Therefore (±π, 0), (±3π, 0), (±5π, 0), (±7π, 0), · · · are saddle points.

21. To find the critical points, we must solve the equations

∂f

∂x= ex(1− cos y) = 0

∂f

∂y= ex(sin y) = 0.

The first equation has solutiony = 0,±2π,±4π, . . . .

The second equation has solutiony = 0,±π,±2π,±3π, . . . .

Since x can be anything, the linesy = 0,±2π,±4π, . . .

are lines of critical points.We calculate

D = (fxx)(fyy)− (fxy)2 = ex(1− cos y)ex cos y − (ex sin y)2

= e2x(cos y − cos2 y − sin2 y)

= e2x(cos y − 1)

At any critical point on one of the lines y = 0, y = ±2π, y = ±4π, . . .,

D = e2x(1− 1) = 0.

Thus, D tells us nothing. However, all along these critical lines, the value of the function, f , is zero. Since the function fis never negative, the critical points are all both local and global minima.

22. At a critical point,

fx = cosx sin y = 0 so cosx = 0 or sin y = 0;

and

fy = sinx cos y = 0 so sinx = 0 or cos y = 0.

Case 1: Assume cosx = 0. This gives

x = · · · − 3π

2,−π

2,π

2,

3π

2, · · ·

(This can be written more compactly as: x = kπ + π/2, for k = 0,±1,±2, · · ·.)If cosx = 0, then sinx = ±1 6= 0. Thus in order to have fy = 0 we need cos y = 0, giving

y = · · · − 3π

2,−π

2,π

2,

3π

2, · · ·


(More compactly, y = lπ + π/2, for l = 0,±1,±2, · · ·)Case 2: Assume sin y = 0. This gives

y = · · · − 2π,−π, 0, π, 2π, · · ·(More compactly, y = lπ, for l = 0,±1,±2, · · ·)If sin y = 0, then cos y = ±1 6= 0, so to get fy = 0 we need sinx = 0, giving

x = · · · ,−2π,−π, 0, π, 2π, · · ·(More compactly, x = kπ for k = 0,±0,±1,±2, · · ·)Hence we get all the critical points of f(x, y). Those from Case 1 are as follows:

· · ·(−π

2,−π

2

),(−π

2,π

2

),(−π

2,

3π

2

)· · ·

· · ·(π

2,−π

2

),(π

2,π

2

),(π

2,

3π

2

)· · ·

· · ·(

3π

2,−π

2

),(

3π

2,π

2

),(

3π

2,

3π

2

)· · ·

Those from Case 2 are as follows:

· · · (−π,−π), (−π, 0), (−π, π), (−π, 2π) · · ·

· · · (0,−π), (0, 0), (0, π), (0, 2π) · · ·· · · (π,−π), (π, 0), (π, π), (π, 2π) · · ·

More compactly these points can be written as,

(kπ, lπ), for k = 0,±1,±2, · · · , l = 0,±1,±2, · · ·and (kπ +

π

2, lπ +

π

2), for k = 0,±1,±2, · · · , l = 0,±1,±2, · · ·

To classify the critical points, we find the discriminant. We have

fxx = − sinx sin y, fyy = − sinx sin y, and fxy = cosx cos y.

Thus the discriminant is


= (− sinx sin y)(− sinx sin y)− (cosx cos y)2

= sin2 x sin2 y − cos2 x cos2 y

= sin2 y − cos2 x. (Use: sin2 x = 1− cos2 x and factor.)

At points of the form (kπ, lπ) where k = 0,±1,±2, · · · ; l = 0,±1,±2, · · ·, we haveD(x, y) = −1 < 0 so (kπ, lπ) are saddle points.At points of the form (kπ + π

2, lπ + π

2) where k = 0,±1,±2, · · · ; l = 0,±1,±2, · · ·

D(kπ + π2, lπ + π

2) = 1 > 0, so we have two cases:

If k and l are both even or k and l are both odd, thenfxx = − sinx sin y = −1 < 0, so (kπ + π

2, lπ + π

2) are local maximum points.

If k is even but l is odd or k is odd but l is even, thenfxx = 1 > 0 so (kπ + π

2, lπ + π

2) are local minimum points.

23. At a local maximum value of f ,∂ f

∂ x= −2x−B = 0.

We are told that this is satisfied by x = −2. So −2(−2)−B = 0 and B = 4. In addition,

∂ f

∂ y= −2y − C = 0

and we know this holds for y = 1, so −2(1) − C = 0, giving C = −2. We are also told that the value of f is 15 at thepoint (−2, 1), so

15 = f(−2, 1) = A− ((−2)2 + 4(−2) + 12 − 2(1)) = A− (−5), so A = 10.

15.1 SOLUTIONS 1045

Now we check that these values of A, B, and C give f(x, y) a local maximum at the point (−2, 1). Since

fxx(−2, 1) = −2,

fyy(−2, 1) = −2

andfxy(−2, 1) = 0,

we have that fxx(−2, 1)fyy(−2, 1) − f2xy(−2, 1) = (−2)(−2) − 0 > 0 and fxx(−2, 1) < 0. Thus, f has a local

maximum value 15 at (−2, 1).

24. (a) (1, 3) is a critical point. Since fxx > 0 and the discriminant

D = fxxfyy − f2xy = fxxfyy − 02 = fxxfyy > 0,

the point (1, 3) is a minimum.(b) See Figure 15.3.

1

3

x

y

0

14

1632

6412

0

Figure 15.3

1

1

x

y

0

0

0

0

13579

13

5 7 9−1

−3−5−7

−1

−3

−5

−7

Figure 15.4

25. (a) (a, b) is a critical point. Since the discriminant D = fxxfyy − f2xy = −f2

xy < 0, (a, b) is a saddle point.(b) See Figure 15.4.

26. Begin by constructing little pieces of the contour diagram around each of the points (−1, 0), (3, 3), and (3,−3) wheresome information about f is given. The general shape will be as in Figure 15.5, and the directions of increasing contourvalues are indicated for each part. Then complete the diagram in any way. One possible solution is given in Figure 15.6.

Figure 15.5: Part of contourdiagram with arrows showing

direction of increasingfunction values

01

2

01

2

3

4

Figure 15.6: Contour diagramof f(x, y)


27. Since (2, 4) is a local minimum, the contours around (2, 4) are closed curves with increasing values as we go away fromthe point (2, 4). We assume that the function values continue to increase as we move parallel to the y-axis to the point(2, 1). Since (2, 1) is a saddle point, we draw the contours so that the values go down as we move up or down from thispoint, and up as we move to the left or right. One possible contour diagram is shown in Figure 15.7.

−1 1 2 3 4 5 6

−1

1

2

3

4

5

6

0 1 2 3 4

−1

−1

−2−3

−4

x

y

Figure 15.7

f=−1

f=−1

f=

0

f=

0

f=

1f=

1

(0, 0)

x

y

Figure 15.8

28. (a) Setting the partial derivatives equal to 0, we have

fx(x, y) = 2x(x2 + y) + 2x(x2 − y) = 4x3 = 0

fy(x, y) = −(x2 + y) + (x2 − y) = −2y = 0.

Thus, (0, 0) is the only critical point.(b) Calculating D gives

D = (fxx)(fyy)− (fxy)2 = (12x2)(−2)− 02 = −24x2.

At x = 0, y = 0, we haveD(0, 0) = 0.

Thus the second derivative test tells us nothing about the nature of the critical point.(c) Since f(0, 0) = 0, we sketch contours with values near 0. The contour f = 0 is given by

(x2 − y)(x2 + y) = 0,

that is, the two parabolasy = x2 and y = −x2

We also sketch the contours f = 1 and f = −1. See Figure 15.8.Since there are values of the function which are both positive (above f(0, 0)) and negative (below (f(0, 0)),

near the critical point (0, 0), the origin is neither a local maximum nor a local minimum; it is a saddle point.

29. We have fx = 2kx− 4y and fy = 2y − 4x, so fxx = 2k, fxy = −4, and fyy = 2. The discriminant is

D = (fxx)(fyy)− (fxy)2 = (2k)(2)− (−4)2 = 4k − 16.

Since D = 4k − 16, we see that D < 0 when k < 4. The function has a saddle point at the point (0, 0) when k < 4.When k > 4, we have D > 0 and fxx > 0, so the function has a local minimum at the point (0, 0). When k = 4,the discriminant is zero, and we get no information about this critical point. By looking at the values of the function inTable 15.1, it appears that f has a local minimum at the point (0, 0) when k = 4.

Table 15.1

y

x

−0.1 0 0.1

−0.1 0.01 0.01 0.09

0 0.04 0 0.04

0.1 0.09 0.01 0.01

(a) The function f(x, y) has a saddle point at (0, 0) if k < 4.

15.1 SOLUTIONS 1047

(b) There are no values of k for which this function has a local maximum at the point (0, 0).(c) The function f(x, y) has a local minimum at (0, 0) if k ≥ 4.

30. The first order partial derivatives are

fx(x, y) = 2kx− 2y and fy(x, y) = 2ky − 2x.

And the second order partial derivatives are

fxx(x, y) = 2k fxy(x, y) = −2 fyy(x, y) = 2k

Since fx(0, 0) = fy(0, 0) = 0, the point (0, 0) is a critical point. The discriminant is

D = (2k)(2k)− 4 = 4(k2 − 1).

For k = ±2, the discriminant is positive, D = 12. When k = 2, fxx(0, 0) = 4 which is positive so we have a localminimum at the origin. When k = −2, fxx(0, 0) = −4 so we have a local maximum at the origin. In the case k = 0,D = −4 so the origin is a saddle point.

Lastly, when k = ±1 the discriminant is zero, so the second derivative test can tell us nothing. Luckily, we can factorf(x, y) when k = ±1. When k = 1,

f(x, y) = x2 − 2xy + y2 = (x− y)2.

This is always greater than or equal to zero. So f(0, 0) = 0 is a minimum and the surface is a trough-shaped paraboliccylinder with its base along the line x = y.

When k = −1,f(x, y) = −x2 − 2xy − y2 = −(x+ y)2.

This is always less than or equal to zero. So f(0, 0) = 0 is a maximum. The surface is a parabolic cylinder, with its topridge along the line x = −y.

−1

−4

−8

−12−16

y

x

k = −2

−1

−1

−5

−5

−10

−10

−20

−20

−30

−30

k = −1

x

y k = 0

−1

−1

−4

−4

−8

−8

−12

−12

−16

−16

4

4

8

8

12

12

16

16

1

1

x

y

1

1

5

5

10

10

20

20

30

30

k = 1

x

y

1

4

8

1216

y

x

k = 2

Figure 15.9


31. The partial derivatives are

fx(x, y) = 3x2 − 3y2 and fy(x, y) = −6xy.

Now fx(x, y) will vanish if x = ±y and fy(x, y) will vanish if either x = 0 or y = 0. Since the partial derivativesare defined everywhere, the only critical points are where fx(x, y) and fy(x, y) vanish simultaneously. (0, 0) is the onlycritical point.

To find the contour for f(x, y) = 0, we solve the equation x3 − 3xy2 = 0. This can be factored into

f(x, y) = x(x−√

3 y)(x+√

3 y) = 0

whose roots are x = 0, x =√

3 y and x = −√

3 y. Each of these roots describes a line through the origin; the three ofthem divide the plane into six regions. Crossing any one of these lines will change the sign of only one of the three factorsof f(x, y), which will change the sign of f(x, y).

−2−1 21

0

0

−1−2

0

112

0

−1−2

0

112

� y = x√

3

� y = −x√

3

f > 0f < 0

f < 0f > 0

f > 0 f < 0

x

y

Figure 15.10


Exercises

1. Mississippi lies entirely within a region designated as 80s so we expect both the maximum and minimum daily hightemperatures within the state to be in the 80s. The southwestern-most corner of the state is close to a region designated as90s, so we would expect the temperature here to be in the high 80s, say 87-88. The northern-most portion of the state islocated near the center of the 80s region. We might expect the high temperature there to be between 83-87.

Alabama also lies completely within a region designated as 80s so both the high and low daily high temperatureswithin the state are in the 80s. The southeastern tip of the state is close to a 90s region so we would expect the temperaturehere to be about 88-89 degrees. The northern-most part of the state is near the center of the 80s region so the temperaturethere is 83-87 degrees.

Pennsylvania is also in the 80s region, but it is touched by the boundary line between the 80s and a 70s region. Thuswe expect the low daily high temperature to occur there and be about 80 degrees. The state is also touched by a boundaryline of a 90s region so the high will occur there and be 89-90 degrees.

New York is split by a boundary between an 80s and a 70s region, so the northern portion of the state is likely to beabout 74-76 while the southern portion is likely to be in the low 80s, maybe 81-84 or so.

California contains many different zones. The northern coastal areas will probably have the daily high as low as65-68, although without another contour on that side, it is difficult to judge how quickly the temperature is dropping offto the west. The tip of Southern California is in a 100s region, so there we expect the daily high to be 100-101.

Arizona will have a low daily high around 85-87 in the northwest corner and a high in the 100s, perhaps 102-107 inits southern regions.

Massachusetts will probably have a high daily high around 81-84 and a low daily high of 70.

15.2 SOLUTIONS 1049

2. Let the line be in the form y = b+mx. When x equals−1, 0 and 1, then y equals b−m, b, and b+m, respectively. Thesum of the squares of the vertical distances, which is what we want to minimize, is

f(m, b) = (2− (b−m))2 + (−1− b)2 + (1− (b+m))2.

To find the critical points, we compute the partial derivatives with respect to m and b,

fm = 2(2− b+m) + 0 + 2(1− b−m)(−1)

= 4− 2b+ 2m− 2 + 2b+ 2m

= 2 + 4m,

fb = 2(2− b+m)(−1) + 2(−1− b)(−1) + 2(1− b−m)(−1)

= −4 + 2b− 2m+ 2 + 2b− 2 + 2b+ 2m

= −4 + 6b.

Setting both partial derivatives equal to zero, we get a system of equations:

2 + 4m = 0,

−4 + 6b = 0.

The solution is m = −1/2 and b = 2/3. One can check that it is a minimum. Hence, the regression line is y = 23− 1

2x.

3. The function f has no global maximum or global minimum.

4. The function g has a global minimum (it is 0) but no global maximum.

5. The function h has no global maximum or minimum.

6. Since f(x, y) ≤ 0 for all x, y and sincef(0, 0) = 0, the function has a global maximum (it is 0) and no global minimum.

7. Suppose x is fixed. Then for large values of y the sign of f is determined by the highest power of y, namely y3. Thus,

f(x, y)→∞ as y →∞f(x, y)→ −∞ as y → −∞.

So f does not have a global maximum or minimum.

8. To maximize z = x2 + y2, it suffices to maximize x2 and y2. We can maximize both of these at the same time bytaking the point (1, 1), where z = 2. It occurs on the boundary of the square. (Note: We also have maxima at the points(−1,−1), (−1, 1) and (1,−1) which are on the boundary of the square.)

To minimize z = x2 +y2, we choose the point (0, 0), where z = 0. It does not occur on the boundary of the square.

9. To maximize z = −x2 − y2 it suffices to minimize x2 and y2. Thus, the maximum is at (0, 0), where z = 0. It does notoccur on the boundary of the square.

To minimize z = −x2−y2, it suffices to maximize x2 and y2. Do this by taking the point (1, 1), (−1,−1), (−1, 1),or (1,−1) where z = −2. These occur on the boundary of the square.

10. To maximize this function, it suffices to maximize x2 and minimize y2. We can do this by choosing the point (1, 0), or(−1, 0) where z = 1. These occur on the boundary of the square.

To minimize z = x2 − y2, it suffices to maximize y2 and minimize x2. We can do this by taking the point (0, 1), or(0,−1) where z = −1. These occur on the boundary of the square.

11. The maximum value, which is slightly above 30, say 30.5, occurs approximately at the origin. The minimum value, whichis about 20.5, occurs at (2.5, 5).

12. The maxima occur at about (π/2, 0) and (π/2, 2π). The minimum occurs at (π/2, π). The maximum value is about 1,the minimum value is about −1.

13. The maximum value, which is about 11, occurs at (5.1, 4.9). The minimum value, which is about −1, occurs at (1, 3.9).

Problems

14. To find critical points of g we solve

gx(x, y) = Ax+By −D = 0

gy(x, y) = Bx+ Cy − E = 0


which is equivalent to

Ax+By = D

Bx+ Cy = E.

The second derivative test shows that the critical point gives a minimum for g because gxxgyy − g2xy = AC − B2 > 0

and gxx = A > 0.

15. The variables are a and b, so we set

∂S

∂a= 2(a+ b) + 8(4a+ b− 2) + 18(9a+ b− 4) = 0

∂S

∂b= 2(a+ b) + 2(4a+ b− 2) + 2(9a+ b− 4) = 0,

so, collecting terms and dividing by 4 and 2 respectively,

49a+ 7b− 22 = 0

14a+ 3b− 6 = 0.

Solving gives a = 24/49, b = −2/7.Since there is only one critical point and S is unbounded as a, b → ∞, this critical point is the global minimum.

Therefore, the best fitting parabola is

y =24

49x2 − 2

7.

16. We must find b and m to minimize the function

g(b,m) =

∫ 1

0

(x2 − (b+mx))2dx

=

∫ 1

0

(x4 − 2bx2 + b2 − 2mx3 + 2bmx+m2x2)dx

=1

5− 2

3b+ b2 − m

2+ bm+

m2

3.

The critical points of g are given by the equations

∂g

∂b= −2

3+ 2b+m = 0

∂g

∂m= −1

2+ b+

2

3m = 0.

The solution is b = −1/6 and m = 1. The linear least squares approximation of x2 on the interval [0, 1] is L(x, y) =−1/6 + x.

17. We must find b and m to minimize the function

g(b,m) =

∫ 1

0

(x3 − (b+mx))2dx

=

∫ 1

0

(x6 − 2bx3 + b2 − 2mx4 + 2bmx+m2x2)dx

=1

6− 1

2b+ b2 − 2m

5+ bm+

m2

3.

The critical points of g are given by the equations

∂g

∂b= −1

2+ 2b+m = 0

∂g

∂m= −2

5+ b+

2

3m = 0.

The solution is b = −1/5 and m = 9/10. The linear least squares approximation of x3 on the interval [0, 1] is L(x, y) =−0.2 + 0.9x.

15.2 SOLUTIONS 1051

18. We calculate the partial derivatives and set them to zero.

∂ (range)∂t

= −10t− 6h+ 400 = 0

∂ (range)∂h

= −6t− 6h+ 300 = 0.

10t+ 6h = 400

6t+ 6h = 300

solving we obtain4t = 100

sot = 25

Solving for h, we obtain 6h = 150, yielding h = 25. Since the range is quadratic in h and t, the second derivative testtells us this is a local and global maximum. So the optimal conditions are h = 25% humidity and t = 25◦C.

19. (a) This tells us that an increase in the price of either product causes a decrease in the quantity demanded of bothproducts. An example of products with this relationship is tennis rackets and tennis balls. An increase in the price ofeither product is likely to lead to a decrease in the quantity demanded of both products as they are used together. Ineconomics, it is rare for the quantity demanded of a product to increase if its price increases, so for q1, the coefficientof p1 is negative as expected. The coefficient of p2 in the expression could be either negative or positive. In this case,it is negative showing that the two products are complementary in use. If it were positive, however, it would indicatethat the two products are competitive in use, for example Coke and Pepsi.

(b) The revenue from the first product would be q1p1 = 150p1 − 2p21 − p1p2, and the revenue from the second product

would be q2p2 = 200p2 − p1p2 − 3p22. The total sales revenue of both products, R, would be

R(p1, p2) = 150p1 + 200p2 − 2p1p2 − 2p21 − 3p2

2.

Note that R is a function of p1 and p2. To find the critical points of R, set∇R = 0, i.e.,

∂R

∂p1=∂R

∂p2= 0.

This gives∂R

∂p1= 150− 2p2 − 4p1 = 0

and∂R

∂p2= 200− 2p1 − 6p2 = 0

Solving simultaneously, we have p1 = 25 and p2 = 25. Therefore the point (25, 25) is a critical point forR. Further,

∂2R

∂p21

= −4,∂2R

∂p22

= −6,∂2R

∂p1∂p2= −2,

so the discriminant at this critical point is

D = (−4)(−6)− (−2)2 = 20.

Since D > 0 and ∂2R/∂p21 < 0, this critical point is a local maximum. Since R is quadratic in p1 and p2, this is a

global maximum. Therefore the maximum possible revenue is

R = 150(25) + 200(25)− 2(25)(25)− 2(25)2 − 3(25)2

= (6)(25)2 + 8(25)2 − 7(25)2

= 4375.

This is obtained when p1 = p2 = 25. Note that at these prices, q1 = 75 units, and q2 = 100 units.


20. The total revenue isR = pq = (60− 0.04q)q = 60q − 0.04q2,

and as q = q1 + q2, this gives

R = 60q1 + 60q2 − 0.04q21 − 0.08q1q2 − 0.04q2

2 .

Therefore, the profit is

P (q1, q2) = R− C1 − C2

= −13.7 + 60q1 + 60q2 − 0.07q21 − 0.08q2

2 − 0.08q1q2.

At a local maximum point, we have gradP = ~0 :

∂P

∂q1= 60− 0.14q1 − 0.08q2 = 0,

∂P

∂q2= 60− 0.16q2 − 0.08q1 = 0.

Solving these equations, we find thatq1 = 300 and q2 = 225.

To see whether or not we have found a local maximum, we compute the second-order partial derivatives:

∂2P

∂q21

= −0.14,∂2P

∂q22

= −0.16,∂2P

∂q1∂q2= −0.08.

Therefore,

D =∂2P

∂q21

∂2P

∂q22

− ∂2P

∂q1∂q2= (−0.14)(−0.16)− (−0.08)2 = 0.016,

and so we have found a local maximum point. The graph of P (q1, q2) has the shape of an upside down paraboloid sinceP is quadratic in q1 and q2, hence (300, 225) is a global maximum point.

21. (a) The revenue R = p1q1 + p2q2. Profit = P = R− C = p1q1 + p2q2 − 2q21 − 2q2

2 − 10.

∂P

∂q1= p1 − 4q1 = 0 gives q1 =

p1

4

∂P

∂q2= p2 − 4q2 = 0 gives q2 =

p2

4

Since ∂2P∂q2

1

= −4, ∂2P∂q2

2

= −4 and ∂2P∂q1∂q2

= 0, at (p1/4, p2/4) we have that the discriminant, D = (−4)(−4) > 0

and ∂2P∂q2

1

< 0, thus P has a local maximum value at (q1, q2) = (p1/4, p2/4). Since P is quadratic in q1 and q2, this

is a global maximum. So P =p214

+p224− 2

p21

16− 2

p22

16− 10 =

p218

+p228− 10 is the maximum profit.

(b) The rate of change of the maximum profit as p1 increases is

∂

∂p1(max P ) =

2p1

8=p1

4.

22. We want to maximize the theater’s profit, P , as a function of the two variables (prices) pc and pa. As always, P = R−C,where R is the revenue, R = qcpc + qapa, and C is the cost, which is of the form C = k(qc + qa) for some constant k.Thus,

P (pc, pa) = qcpc + qapa − k(qc + qa)

= rp−3c − krp−4

c + sp−1a − ksp−2

a

To find the critical points, solve∂P

∂pc= −3rp−4

c + 4krp−5c = 0

∂P

∂pa= −sp−2

a + 2ksp−3a = 0.

We get pc = 4k/3 and pa = 2k.

15.2 SOLUTIONS 1053

This critical point is a global maximum by the following useful, general argument. Suppose that F (x, y) = f(x) +g(y), where f has a global maximum at x = b and g has a global maximum at y = d. Then for all x, y:

F (x, y) = f(x) + g(y) ≤ f(b) + g(d) = F (b, d),

so F has global maximum at x = b, y = d.The profit function in this problem has the form

P (pc, pa) = f(pc) + g(pa),

and the usual single-variable calculus argument using f ′ and g′ shows that pc = 4k/3 and pa = 2k are global maximafor f and g, respectively. Thus the maximum profit occurs when pc = 4k/3 and pa = 2k. Thus,

pcpa

=4k/3

2k=

2

3.

23. Let the sides be x, y, z cm. Then the volume is given by V = xyz = 32.The surface area S is given by

S = 2xy + 2xz + 2yz.

Substituting z = 32/(xy) gives

S = 2xy +64

y+

64

x.

At a critical point,

∂S

∂x= 2y − 64

x2= 0

∂S

∂y= 2x− 64

y2= 0,

The symmetry of the equations (or by dividing the equations) tells us that x = y and

2x− 64

x2= 0

x3 = 32

x = 321/3 = 3.17 cm.

Thus the only critical point is x = y = (32)1/3 cm and z = 32/((32)1/3 · (32)1/3

)= (32)1/3 cm. At the critical point

SxxSyy − (Sxy)2 =128

x3· 128

y3− 22 =

(128)2

x3y3− 4.

Since D > 0 and Sxx > 0 at this critical point, the critical point x = y = z = (32)1/3 is a local minimum. SinceS →∞ as x, y →∞, the local minimum is a global minimum.

24. Let the sides of the base be x and y cm. Let the height be z cm. Then the volume is given by xyz = 32 and the surfacearea, S, is given by

S = xy + 2xz + 2yz.

Substituting z = 32/(xy) gives

S = xy +64

y+

64

x.

At a critical point

∂S

∂x= y − 64

x2= 0

∂S

∂y= x− 64

y2= 0.

The symmetry of the equations tells us that x = y and

x− 64

x2= 0

x3 = 64

x = 4 cm.


Thus the only critical point is x = y = 4 cm and z = 32/(4 · 4) = 2 cm. At the critical point

D = SxxSyy − (Sxy)2 =128

x3· 128

y3− 12 =

(128)2

x3y3− 1.

Since D > 0 and Sxx > 0 at this critical point, the critical point x = y = 4, z = 2 is a local minimum. Since S →∞ asx, y →∞, the local minimum is a global minimum.

25. If the coordinates of the corner on the plane are (x, y, z), the volume of the box is V = xyz. Since z = 1− 3x− 2y onthe plane, the volume is given by

V = xy(1− 3x− 2y) = xy − 3x2y − 2xy2.

The domain is the triangular region 0 ≤ x ≤ 13, 0 ≤ y ≤ (1− 3x)/2. At a critical point,

∂V

∂x= y − 6xy − 2y2 = y(1− 6x− 2y) = 0

∂V

∂y= x− 3x2 − 4xy = x(1− 3x− 4y) = 0,

One solution is x = y = 0. Another is x = 0, y = 12

; another is y = 0, x = 13

. Another is the solution of

1− 6x− 2y = 0

1− 3x− 4y = 0,

namely x = 19, y = 1

6.

If either x = 0 or y = 0, then V = 0, so these solutions do not give the maximum volume. Since

D = VxxVyy − (Vxy)2 = (−6y)(−4x)− (1− 6x− 4y)2

D(

1

9,

1

6

)=(−6 · 1

6

)(−4 · 1

9

)−(

1− 6 · 1

9− 4 · 1

6

)2

=4

9− 1

9=

1

3> 0,

and Vxx( 19, 1

6) = −1 < 0, the point x = 1

9, y = 1

6, is a local maximum at which V = (1/9)(1/6) − 3(1/9)2(1/6) −

2(1/9)(1/6)2 = 1/162.Since all points on the boundary of the domain give V = 0, the local maximum is a global maximum.

26.

wl

h

Figure 15.11

Let w, h and l be width, height and length of the suitcase in cm. Then its volume V = lwh, and w + h+ l ≤ 135.To maximize the volume V , choose w + h+ l = 135, and thus l = 135− w − h,

V = wh(135− w − h)

= 135wh− w2h− wh2

Differentiating gives

Vw = 135h− 2wh− h2,

Vh = 135w − w2 − 2wh.

Find the critical points by solving Vw = 0 and Vh = 0:

Vw = 0 gives 135h− h2 = 2wh,

Vh = 0 gives 135w − w2 = 2wh.

15.2 SOLUTIONS 1055

As hw 6= 0, we cancel h (and w respectively) in the above equations and get

135− h = 2w

135− w = 2h

Subtracting givesw − h = 2(w − h)

hence w = h. Therefore, substituting into the equation Vw = 0

135h− h2 = 2h2

and therefore3h2 = 135h.

Since h 6= 0, we have

h =135

3= 45.

So w = h = 45 cm. Thus, l = 135− w − h = 45 cm. To check that this critical point is a maximum, we find

Vww = −2h, Vhh = −2w,

Vwh = 135− 2w − 2h,

soD = VwwVhh − V 2

wh = 4hw − (135− 2w − 2h)2.

At w = h = 45, we have Vww = −2(45) < 0 and D = 4(45)2 − (135− 90− 90)2 = 6075 > 0, hence V is maximumat w = h = l = 45.

Therefore, the suitcase with maximum volume is a cube with dimensions width = height = length = 45 cm.

27.

w

l

h

Figure 15.12

The box is shown in Figure 15.12. Cost of four sides = (2hl + 2wh)(1)c/. Cost of two bottoms = (2wl)(2)c/. Thusthe total cost C (in cents) of the box is

C = 2(hl + wh) + 4wl.

But volume wlh = 512, so l = 512/(wh), thus

C =1024

w+ 2wh+

2048

h.

To minimize C, find the critical points of C by solving

Ch = 2w − 2048

h2= 0,

Cw = 2h− 1024

w2= 0.

We get

2wh2 = 2048

2hw2 = 1024.


Since w, h 6= 0, we can divide the first equation by the second giving

2wh2

2hw2=

2048

1024,

soh

w= 2,

thush = 2w.

Substituting this in Ch = 0, we obtain h3 = 2048, so h = 12.7 cm. Thus w = h/2 = 6.35 cm, and l = 512/(wh) =6.35 cm. Now we check that these dimensions minimize the cost C. We find that

D = ChhCww − C2hw = (

4096

h3)(

2048

w3)− 22,

and at h = 12.7, w = 6.35, Chh > 0 and D = 16 − 4 > 0, thus C has a local minimum at h = 12.7 and w = 6.35.Since C increases without bound as w, h→ 0 or∞, this local minimum must be a global minimum.

Therefore, the dimensions of the box that minimize the cost are w = 6.35 cm, l = 6.35 cm and h = 12.7 cm.

28. The square of the distance from the point (x, y, z) to the origin is

S = x2 + y2 + z2.

If the point is on the plane, z = 1− 3x− 2y, we have

S = x2 + y2 + (1− 3x− 2y)2.

At the critical point

∂S

∂x= 2x+ 2(1− 3x− 2y)(−3) = 2(10x+ 6y − 3) = 0

∂S

∂y= 2y + 2(1− 3x− 2y)(−2) = 2(6x+ 5y − 2) = 0.

Simplifying gives10x+ 6y = 3

6x+ 5y = 2,

with solution x = 3/14, y = 1/7. At this point z = 1/14. We have

D = SxxSyy − (Sxy)2 = (20)(10)− 122 = 56,

so D > 0 and Sxx > 0. Thus, the point x = 3/14, y = 1/7 is a local minimum. Since S →∞ as x, y → ±∞, the localminimum is a global minimum. Thus, x = 3/14, y = 1/7, z = 1/14 is the closest point to the origin on the plane.

29. We minimize the square of the distance from the point (x, y, z) to the origin:

S = x2 + y2 + z2.

Since z2 = 9− xy − 3x, we haveS = x2 + y2 + 9− xy − 3x.

At a critical point

∂S

∂x= 2x− y − 3 = 0

∂S

∂y= 2y − x = 0,

so x = 2y, and2(2y)− y − 3 = 0

giving y = 1, so x = 2 and z2 = 9− 2 · 1− 3 · 2 = 1, so z = ±1. We have

D = SxxSyy − (Sxy)2 = 2 · 2− (−1)2 = 4− 1 > 0,

so, since D > 0 and Sxx > 0, the critical points are local minima. Since S → ∞ as x, y → ±∞, the local minima areglobal minima.

If x = 2, y = 1, z = ±1, we have S = 22 + 12 + 12 = 6, so the shortest distance to the origin is√

6.

15.2 SOLUTIONS 1057

30. (a) Let t be the number of years since 1960 and let P (t) be the population in millions in the year 1960 + t. We assumethat P = Ceat, and therefore

lnP = at+ lnC.

So, we plot lnP against t and find the line of best fit. Our data points are (0, ln 180), (10, ln 206), and (20, ln 226).Applying the method of least squares to find the best-fitting line, we find that

a =ln 226− ln 180

20≈ 0.0114,

lnC =ln 206

3− ln 226

6+

5 ln 180

6≈ 5.20

Then, C = e5.20 = 181.3 and soP (t) = 181.3e0.0114t.

In 1990, we have t = 30 and the predicted population in millions is

P (30) = 181.3e0.01141(30) = 255.3.

(b) The difference between the actual and the predicted population is about 6 million or 2 12%. Given that only three data

points were used to calculate a and c, this discrepancy is not surprising. Thus, the 1990 census data does not meanthat the assumption of exponential growth is unjustified.

(c) In 2010, we have t = 50 and P (50) = 320.7.

31. Let P (K,L) be the profit obtained using K units of capital and L units of labor. The cost of production is given by

C(K,L) = kK + `L,

and the revenue function is given byR(K,L) = pQ = pAKaLb.

Hence, the profit isP = R− C = pAKaLb − (kK + `L).

In order to find local maxima of P , we calculate the partial derivatives and see where they are zero. We have:

∂P

∂K= apAKa−1Lb − k,

∂P

∂L= bpAKaLb−1 − `.

The critical points of the function P (K,L) are solutions (K,L) of the simultaneous equations:

k

a= pAKa−1Lb,

`

b= pAKaLb−1.

Multiplying the first equation by K and the second by L, we get

kK

a=`L

b,

and soK =

à

kbL.

Substituting for K in the equation k/a = pAKa−1Lb, we get:

k

a= pA

(à

kb

)a−1

La−1Lb.

We must therefore have

L1−a−b = pA(a

k

)a ( `b

)a−1

.

Hence, if a+ b 6= 1,

L =

[pA(a

k

)a ( `b

)(a−1)]1/(1−a−b)

,


and

K =à

kbL =

à

kb

[pA(a

k

)a ( `b

)(a−1)]1/(1−a−b)

.

To see if this is really a local maximum, we apply the second derivative test. We have:

∂2P

∂K2= a(a− 1)pAKa−2Lb,

∂2P

∂L2= b(b− 1)pAKaLb−2,

∂2P

∂K∂L= abpAKa−1Lb−1.

Hence,

D =∂2P

∂K2

∂2P

∂L2−(

∂2P

∂K∂L

)2

= ab(a− 1)(b− 1)p2A2K2a−2L2b−2 − a2b2p2A2K2a−2L2b−2

= ab((a− 1)(b− 1)− ab)p2A2K2a−2L2b−2

= ab(1− a− b)p2A2K2a−2L2b−2.

Now a, b, p, A, K, and L are positive numbers. So, the sign of this last expression is determined by the sign of 1− a− b.(a) We assumed that a + b < 1, so D > 0, and as 0 < a < 1, then ∂2P/∂K2 < 0 and so we have a unique local

maximum. To verify that the local maximum is a global maximum, we focus on the cost. Let C = kK + `L. SinceK ≥ 0 and L ≥ 0, K ≤ C/k and L ≤ C/`. Therefore the profit satisfies:

P = pAKaLb − (kK + `L)

≤ pA(C

k

)a (C`

)b− C

= mCa+b − C

where m = pA(1/k)a(1/`)b. Since a+ b < 1, the profit is negative for large costs C, say C ≥ C0 (C0 = m1−a−b

will do). Therefore, in the KL-plane for K ≥ 0 and L ≥ 0, the profit is less than or equal to zero everywhere on orabove the line kK+ `L = Co. Thus the global maximum must occur inside the triangle bounded by this line and theK and L axes. Since P ≤ 0 on the K and L axes as well, the global maximum must be in the interior of the triangleat the unique local maximum we found.

In the case a + b < 1, we have decreasing returns to scale. That is, if the amount of capital and labor used ismultiplied by a constant λ > 0, we get less than λ times the production.

(b) Now suppose a+ b ≥ 1. If we multiply K and L by λ for some λ > 0, then

Q(λK, λL) = A(λK)a(λL)b = λa+bQ(K,L).

We also see thatC(λK, λL) = λC(K,L).

So if a+ b = 1, we haveP (λK, λL) = λP (K,L).

Thus, if λ = 2, so we are doubling the inputs K and L, then the profit P is doubled and hence there can be nomaximum profit.

If a+ b > 1, we have increasing returns to scale and there can again be no maximum profit: doubling the inputswill more than double the profit. In this case, the profit increases without bound as K, L go toward infinity.

32. We have

fx = 2x(y + 1)3 = 0 only when x = 0 or y = −1

fy = 3x2(y + 1)2 + 2y = 0 never when y = −1 and only for y = 0 when x = 0

We conclude that fx = 0 and fy = 0 only when x = 0, y = 0, so f has only one critical point, namely (0, 0).

15.2 SOLUTIONS 1059

The second derivative test at (0, 0) gives

D = fxxfyy − (fxy)2 = 2(y + 1)3(6x2(y + 1) + 2)− (6x(y + 1)2)2

= 2(1)(2)− 0 > 0 when x = 0, y = 0

Since fxx > 0 at (0, 0), this means f has a local minimum at (0, 0).[Alternatively, if we expand (y + 1)3, then we can view f(x, y) as x2 + y2+ (terms of degree 3 or greater in x and

y), which means that f behaves likes x2 + y2 near (0, 0).]Although (0, 0) is a local minimum, it cannot be a global minimum since for fixed x, say x = 1, the function f(x, y)

is a cubic polynomial in y and cubics take on arbitrarily large positive and negative values.In the single-variable case, suppose a function f defined on the real line is differentiable and its derivative is con-

tinuous. Then if f has only one critical point, say x = 0, then if that critical point is a local minimum, it must also be aglobal minimum. This is because f ′ cannot change sign without f ′ = 0 so we must have f ′ < 0 for x < 0 and f ′ > 0for x > 0. Thus f is decreasing for all x < 0 and increasing for all x > 0, which makes x = 0 the global minimum forf .

33. (a) We have f(2, 1) = 120.(i) If x > 20 then f(x, y) > 10x > 200 > f(2, 1).

(ii) If y > 20 then f(x, y) > 20y > 400 > f(2, 1).(iii) If x < 0.01 and y ≤ 20 then f(x, y) > 80/(xy) > 80/((0.01)(20)) = 400 > f(2, 1).(iv) If y < 0.01 and x ≤ 20 then f(x, y) > 80/(xy) > 80/((20)(0.01)) = 400 > f(2, 1).

(b) The continuous function f must achieve a minimum at some point (x0, y0) in the closed and bounded region R′ :0.01 ≤ x ≤ 20, 0.01 ≤ y ≤ 20. Since (2, 1) is in R′, we must have f(x0, y0) ≤ f(2, 1). By part (a), f(x0, y0)is less than all values of f in the part of R that is outside R′, so f(x0, y0) is a minimum for f on all of R. Since(x0, y0) is not on the boundary of R, it must be a critical point of f .

(c) The only critical point of f in R is the point (2, 1), so by part (b) f has a global minimum there.

34. (a) The function f is continuous in the region R, but R is not closed and bounded so a special analysis is required.Notice that f(x, y) tends to∞ as (x, y) tends farther and farther from the origin or tends toward any point on

the x or y axis. This suggests that a minimum for f , if it exists, can not be too far from the origin or too close to theaxes. For example, if x > 10 then f(x, y) > 4x > 40, and if y > 10 then f(x, y) > 5y > 50. If 0 < x < 0.1 thenf(x, y) > 2/x > 20, and if 0 < y < 0.1 then f(x, y) > 3/y > 30.

Since f(1, 1) = 14, a global minimum for f if it exists must be in the smaller region R′ : 0.1 ≤ x ≤ 10,0.1 ≤ y ≤ 10. The region R′ is closed and bounded and so f does have a minimum value at some point in R′, andsince that value is at most 14, it is also a global minimum for all of R.

(b) Since the region R has no boundary, the minimum value must occur at a critical point of f . At a critical point wehave

fx = − 2

x2+ 4 = 0 fy = − 3

y2+ 5 = 0.

The only critical point is (√

1/2,√

3/5) ≈ (0.7071, 0.7746), at which f achieves the minimum valuef(√

1/2,√

3/5) = 4√

2 + 2√

15 ≈ 13.403.

35. (a) By the chain rule applied to M(a) = f(h(a), a) we have M ′(a) = fx(h(a), a)h′(a) + fa(h(a), a)da/da =fx(h(a), a)h′(a) + fa(h(a), a). Since x = h(a) is a critical point of g(x) = f(x, a), we have g′(h(a)) =fx(h(a), a) = 0. Thus M ′(a) = fa(h(a), a).

(b) Let n(a) = f(x, a) be a cross-section of f with x fixed. We have n′(a) = fa(x, a). If x = h(a), then n(a) =f(h(a), a) = M(a) so the point (a,M(a)) is on the graph of n. And n′(a) = fa(h(a), a) = M ′(a) so n and Mhave the same slope at the point (a, h(a)). Their graphs are tangent.

(c) The quadratic function g(x) = f(x, a) = −(1/2)x2 + 2ax − a2 has a maximum where g′(x) = −x + 2a = 0 orx = 2a. Thus h(a) = 2a. The maximum value is M(a) = g(h(a)) = f(2a, a) = a2. The graph in Figure 15.13shows that the graph of M(a) is the upper boundary, called the upper envelope, of the family of cross-sections of fwith x fixed.


M(a)

f(−2, a)

f(−1, a)

f(−0, a) f(1, a)

f(2, a)

f(3, a)

x

y

Figure 15.13


Exercises

1. Our objective function is f(x, y) = x+ y and our equation of constraint is g(x, y) = x2 + y2 = 1. To optimize f(x, y)with Lagrange multipliers, we solve∇f(x, y) = λ∇g(x, y) subject to g(x, y) = 1. The gradients of f and g are

∇f(x, y) = ~i +~j ,

∇g(x, y) = 2x~i + 2y~j .

So the equation∇f = λ∇g becomes~i +~j = λ(2x~i + 2y~j )

Solving for λ gives

λ =1

2x=

1

2y,

which tells us that x = y. Going back to our equation of constraint, we use the substitution x = y to solve for y:

g(y, y) = y2 + y2 = 1

2y2 = 1

y2 =1

2

y = ±√

1

2= ±√

2

2.

Since x = y, our critical points are (√

22,√

22

) and (−√

22,−√

22

). Since the constraint is closed and bounded, maximumand minimum values of f subject to the constraint exist. Evaluating f at the critical points we find that the maximumvalue is f(

√2

2,√

22

) =√

2 and the minimum value is f(−√

22,−√

22

) = −√

2.

2. Our objective function is f(x, y) = 3x− 2y and our equation of constraint is g(x, y) = x2 + 2y2 = 44. Their gradientsare

∇f(x, y) = 3~i − 2~j ,

∇g(x, y) = 2x~i + 4y~j .

15.3 SOLUTIONS 1061

So the equation∇f = λ∇g becomes 3~i − 2~j = λ(2x~i + 4y~j ). Solving for λ gives us

λ =3

2x=−2

4y,

which we can use to find x in terms of y:

3

2x=−2

4y

−4x = 12y

x = −3y.

Using this relation in our equation of constraint, we can solve for y:

x2 + 2y2 = 44

(−3y)2 + 2y2 = 44

9y2 + 2y2 = 44

11y2 = 44

y2 = 4

y = ±2.

Thus, the critical points are (−6, 2) and (6,−2). Since the constraint is closed and bounded, maximum and minimumvalues of f subject to the constraint exist. Evaluating f at the critical points, we find that the maximum is f(6,−2) =18 + 4 = 22 and the minimum value is f(−6, 2) = −18− 4 = −22.

3. Our objective function is f(x, y) = xy and our equation of constraint is g(x, y) = 4x2 + y2 = 8. Their gradients are

∇f(x, y) = y~i + x~j ,

∇g(x, y) = 8x~i + 2y~j .

So the equation∇f = λ∇g becomes y~i + x~j = λ(8x~i + 2y~j ). This gives

8xλ = y and 2yλ = x.

Multiplying, we get8x2λ = 2y2λ.

If λ = 0, then x = y = 0, which does not satisfy the constraint equation. So λ 6= 0 and we get

2y2 = 8x2

y2 = 4x2

y = ±2x.

To find x, we substitute for y in our equation of constraint.

4x2 + y2 = 8

4x2 + 4x2 = 8

x2 = 1

x = ±1

So our critical points are (1, 2), (1,−2), (−1, 2) and (−1,−2). Since the constraint is closed and bounded, maximumand minimum values of f subject to the constraint exist. Evaluating f(x, y) at the critical points, we have

f(1, 2) = f(−1,−2) = 2

f(1,−2) = f(1,−2) = −2.

Thus, the maximum value of f on g(x, y) = 8 is 2, and the minimum value is −2.


4. The objective function is f(x, y) = x2 + y2 and the equation of constraint is g(x, y) = x4 + y4 = 2. Their gradients are

∇f(x, y) = 2x~i + 2y~j ,

∇g(x, y) = 4x3~i + 4y3~j .

So the equation∇f = λ∇g becomes 2x~i + 2y~j = λ(4x3~i + 4y3~j ). This tells us that

2x = 4λx3,

2y = 4λy3.

Now if x = 0, the first equation is true for any value of λ. In particular, we can choose λ which satisfies the secondequation. Similarly, y = 0 is solution.

Assuming both x 6= 0 and y 6= 0, we can divide to solve for λ and find

λ =2x

4x3=

2y

4y3

1

2x2=

1

2y2

y2 = x2

y = ±x.

Going back to our equation of constraint, we find

g(0, y) = 04 + y4 = 2, so y = ± 4√

2

g(x, 0) = x4 + 04 = 2, so x = ± 4√

2

g(x,±x) = x4 + (±x)4 = 2, so x = ±1.

Thus, the critical points are (0,± 4√

2), (± 4√

2, 0), (1,±1) and (−1,±1). Since the constraint is closed and bounded,maximum and minimum values of f subject to the constraint exist. Evaluating f at the critical points, we find

f(1, 1) = f(1,−1) = f(−1, 1) = f(−1,−1) = 2,

f(0,4√

2) = f(0,− 4√

2) = f(4√

2, 0) = f(− 4√

2, 0) =√

2.

Thus, the minimum value of f(x, y) on g(x, y) = 2 is√

2 and the maximum value is 2.

5. The objective function is f(x, y) = x2 + y2 and the constraint equation is g(x, y) = 4x − 2y = 15, so grad f =(2x)~i + (2y)~j and grad g = 4~i − 2~j . Setting grad f = λ grad g gives

2x = 4λ,

2y = −2λ.

From the first equation we have λ = x/2, and from the second equation we have λ = −y. Setting these equal gives

y = −0.5x.

Substituting this into the constraint equation 4x− 2y = 15 gives x = 3. The only critical point is (3,−1.5).We have f(3,−1.5) = (3)2 + (1.5)2 = 11.25. One way to determine if this point gives a maximum or minimum

value or neither for the given constraint is to examine the contour diagram of f with the constraint sketched in, Fig-ure 15.14. It appears that moving away from the point P = (3,−1.5) in either direction along the constraint increases thevalue of f , so (3,−1.5) is a point of minimum value.

15.3 SOLUTIONS 1063

1 2 3 4

5 10 15 20

−3

−2

−1

1Constraint: 4x− 2y = 15

P = (3,−1.5)

x

y

Figure 15.14

−2 2

−2

2

(√

5/2, 1/2)(−√

5/2,−1/2)

0

0.75

1.5

2.25

x2 − y2 = 1

x

y

Figure 15.15: Graph of x2 − y2 = 1

6. Our objective function is f(x, y) = x2 + y and our equation of constraint is g(x, y) = x2 − y2 = 1. Their gradients are

∇f(x, y) = 2x~i +~j ,

∇g(x, y) = 2x~i − 2y~j .

Thus∇f = λ∇g gives

2x = λ2x

1 = −λ2y

But x cannot be zero, since the constraint equation, −y2 = 1, would then have no real solution for y. So the equation∇f = λ∇g becomes

λ =2x

2x=

1

−2y

1 =1

−2y

−2y = 1

y = −1

2.

Substituting this into our equation of constraint we find

g(x,−1

2) = x2 −

(−1

2

)2

= 1

x2 =5

4

x = ±√

5

2.

So the critical points are (√

52,− 1

2) and (−

√5

2,− 1

2). Evaluating f at these points we find f(

√5

2,− 1

2) = f(−

√5

2,− 1

2) =

54− 1

2= 3

4. This is the minimum value for f(x, y) constrained to g(x, y) = 1. To see this, note that for x2 = y2 + 1,

f(x, y) = y2 + 1 + y = (y + 1/2)2 + 3/4 ≥ 3/4. Alternatively, see Figure 15.15. To see that f has no maximum ong(x, y) = 1, note that f →∞ as x→∞ and y →∞ on the part of the graph of g(x, y) = 1 in quadrant I.

7. The objective function is f(x, y) = x2−xy+ y2 and the equation of constraint is g(x, y) = x2− y2 = 1. The gradientsof f and g are

∇f(x, y) = (2x− y)~i + (−x+ 2y)~j ,

∇g(x, y) = 2x~i − 2y~j .

Therefore the equation∇f(x, y) = λ∇g(x, y) gives

2x− y = 2λx

−x+ 2y = −2λy

x2 − y2 = 1.


Let us suppose that λ = 0. Then 2x = y and 2y = x give x = y = 0. But (0, 0) is not a solution of the third equation,so we conclude that λ 6= 0. Now let’s multiply the first two equations

−2λy(2x− y) = 2λx(−x+ 2y).

As λ 6= 0, we can cancel it in the equation above and after doing the algebra we get

x2 − 4xy + y2 = 0

which gives x = (2 +√

3)y or x = (2−√

3)y.If x = (2 +

√3)y, the third equation gives

(2 +√

3)2y2 − y2 = 1

so y ≈ ±0.278 and x ≈ ±1.038. These give the critical points (1.038, 0.278), (−1.038,−0.278).If x = (2−

√3)y, from the third equation we get

(2−√

3)2y2 − y2 = 1.

But (2−√

3)2 − 1 ≈ −0.928 < 0 so the equation has no solution. Evaluating f gives

f(1.038, 0.278) = f(−1.038,−0.278) ≈ 0.866

Since y →∞ on the constraint, rewriting f as

f(x, y) =(x− y

2

)2

+3

4y2

shows that f has no maximum on the constraint. The minimum value of f is 0.866. See Figure 15.16.

−2

2

(1.038, 0.278)

(−1.038,−0.278) 0.2

0.866

2

x

x2 − y2 = 1

y

Figure 15.16

8. The objective function is f(x, y, z) = x + 3y + 5z and the equation of constraint is g(x, y, z) = x2 + y2 + z2 = 1.Their gradients are

∇f(x, y, z) = ~i + 3~j + 5~k ,

∇g(x, y, z) = 2x~i + 2y~j + 2z~k .

So the equation∇f = λ∇g becomes~i + 3~j + 5~k = λ(2x~i + 2y~j + 2z~k ). Solving for λ we find

λ =1

2x=

3

2y=

5

2z.

Which provides us with the equations

2y = 6x

10x = 2z.

15.3 SOLUTIONS 1065

Solving the first equation for y gives us y = 3x. Solving the second equation for z gives us z = 5x. Substituting theseinto the equation of constraint, we can find x:

x2 + (3x)2 + (5x)2 = 1

x2 + 9x2 + 25x2 = 1

35x2 = 1

x2 =1

35

x = ±√

1

35= ±√

35

35.

Since y = 3x and z = 5x, the critical points are at±(√

3535, 3√

3535,√

357

). Since the constraint is closed and bounded, max-imum and minimum values of f subject to the constraint exist. Evaluating f at the critical points, we find the maximumis f(

√35

35, 3√

3535,√

357

) =√

35 3535

=√

35, and the minimum value is f(−√

3535,−3

√35

35,−√

357

) = −√

35.

9. Our objective function is f(x, y, z) = x2−2y+2z2 and our equation of constraint is g(x, y, z) = x2 +y2 +z2−1 = 0.To optimize f(x, y, z) with Lagrange multipliers, we solve ∇f(x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. Thegradients of f and g are

∇f(x, y, z) = 2x~i − 2~j + 4z~k ,

∇g(x, y) = 2x~i + 2y~j + 2z~k .

We get,

x = λx

−1 = λy

2z = λz

x2 + y2 + z2 = 1.

From the first equation we get x = 0 or λ = 1.If x = 0 we have

−1 = λy

2z = λz

y2 + z2 = 1.

From the second equation z = 0 or λ = 2. So if z = 0, we have y = ±1 and we get the solutions (0, 1, 0),(0,−1, 0). Ifz 6= 0 then λ = 2 and y = − 1

2. So z2 = 3

4which gives the solutions (0,− 1

2,√

32

), (0,− 12,−√

32

).If x 6= 0, then λ = 1, so y = −1, which implies, from the equation x2 + y2 + z2 = 1, that x = 0, which contradicts

the assumption.Since the constraint is closed and bounded, maximum and minimum values of f subject to the constraint exist. There-

fore, evaluating f at the critical points, we get f(0, 1, 0) = −2, f(0,−1, 0) = 2 and f(0,− 12,√

32

) = f(0,− 12,−√

32

) =4. So the maximum value of f is 4 and the minimum is −2.

10. Our objective function is f(x, y, z) = 2x + y + 4z and our equation of constraint is g(x, y, z) = x2 + y + z2 = 16.Their gradients are

∇f(x, y, z) = 2~i + 1~j + 4~k ,

∇g(x, y, z) = 2x~i + 1~j + 2z~k .

So the equation∇f = λ∇g becomes 2~i + 1~j + 4~k = λ(2x~i + 1~j + 2z~k ). Solving for λ we find

λ =2

2x=

1

1=

4

2z

λ =1

x= 1 =

2

z.

Which tells us that x = 1 and z = 2. Going back to our equation of constraint, we can solve for y.

g(1, y, 2) = 16

12 + y + 22 = 16

y = 11.


So our one critical point is at (1, 11, 2). The value of f at this point is f(1, 11, 2) = 2 + 11 + 8 = 21. This is themaximum value of f(x, y, z) on g(x, y, z) = 16. To see this, note that for y = 16− x2 − z2,

f(x, y, z) = 2x+ 16− x2 − z2 + 4z = 21− (x− 1)2 − (z − 2)2 ≤ 21.

As y → −∞, the point (−√16− y, y, 0) is on the constraint and f(−√16− y, y, 0) → −∞, so there is no minimumvalue for f(x, y, z) on g(x, y, z) = 16.

11. The region x2 + y2 ≤ 4 is the shaded disk of radius 2 centered at the origin (including the circle x2 + y2 = 4) shown inFigure 15.17.

We will first find the local maxima and minima in the interior of the disk. So we need to find the extrema of

f(x, y) = x2 + 2y2 in the region x2 + y2 < 4.

For this we compute the critical points:

fx = 2x = 0

fy = 4y = 0

So the critical point is (0, 0). As fxx(0, 0) = 2, fyy(0, 0) = 4 and fxy(0, 0) = 0 we have

D = fxx(0, 0) · fyy(0, 0)− (fxy(0, 0))2 = 8 > 0 and fxx(0, 0) = 2 > 0.

Therefore (0, 0) is a minimum point and f(0, 0) = 0.Now let’s find the local extrema of f on the boundary of the disk, hence this time we have to solve a constraint

problem. We want the extrema of f(x, y) = x2 + 2y2 subject to g(x, y) = x2 + y2 − 4 = 0. We use Lagrangemultipliers:

grad f = λ grad g and x2 + y2 = 4,

which give

2x = 2λx

4y = 2λy

x2 + y2 = 4.

From the first equation we have x = 0 or λ = 1. If x = 0, from the last equation y2 = 4 and therefore (0, 2) and(0,−2) are solutions.

If x 6= 0 then λ = 1 and from the second equation y = 0. Substituting this into the third equation we get x2 = 4 so(2, 0) and (−2, 0) are the other two solutions.

The region x2 + y2 ≤ 4 is closed and bounded, so maximum and minimum values of f in the the region exist.Therefore, as f(0, 2) = f(0,−2) = 8 and f(2, 0) = f(−2, 0) = 4, (0, 2) and (0,−2) are global maxima and (0, 0) isthe global minimum on the whole region. The maximum value of f is 8 and the minimum value of f is 0.

−4 4

−4

4

14

8

16

x

y

Figure 15.17

−√

2√

2

−√

2

√2

1

3

5

−1

−3

−5

x

y

Figure 15.18

15.3 SOLUTIONS 1067

12. The region x2 +y2 ≤ 2 is the shaded disk of radius√

2 centered at the origin (including the circle x2 +y2 = 2) as shownin Figure 15.18.

We first find the local maxima and minima of f in the interior of our disk. So we need to find the extrema of

f(x, y) = x+ 3y, in the region x2 + y2 < 2.

As

fx = 1

fy = 3

f does not have critical points. Now let’s find the local extrema of f on the boundary of the disk. We want to find theextrema of f(x, y) = x+ 3y subject to the constraint g(x, y) = x2 + y2 − 2 = 0. We use Lagrange multipliers


which give

1 = 2λx

3 = 2λy

x2 + y2 = 2.

As λ cannot be zero, we solve for x and y in the first two equations and get x = 12λ

and y = 32λ

. Plugging into the thirdequation gives

8λ2 = 10

so λ = ±√

52

and we get the solutions ( 1√5, 3√

5) and (− 1√

5,− 3√

5). Evaluating f at these points gives

f(1√5,

3√5

) = 2√

5 and

f(− 1√5,− 3√

5) = −2

√5

The region x2 + y2 ≤ 2 is closed and bounded, so maximum and minimum values of f in the region exist. Therefore( 1√

5, 3√

5) is a global maximum of f and (− 1√

5,− 3√

5) is a global minimum of f on the whole region x2 + y2 ≤ 2.

13. The domain x2+2y2 ≤ 1 is the shaded interior of the ellipse x2+2y2 = 1 including the boundary, shown in Figure 15.19.

−1 1

− 1√2

− 1√2

0.10.3

0.5

−0.1−0.3

−0.5

x

y

Figure 15.19

First we want to find the local maxima and minima of f in the interior of the ellipse. So we need to find the extremaof

f(x, y) = xy, in the region x2 + 2y2 < 1.


fx = y = 0 and fy = x = 0.

So there is one critical point, (0, 0). As fxx(0, 0) = 0, fyy(0, 0) = 0 and fxy(0, 0) = 1 we have

D = fxx(0, 0) · fyy(0, 0)− (fxy(0, 0))2 = −1 < 0


so (0, 0) is a saddle and f does not have local extrema in the interior of the ellipse.Now let’s find the local extrema of f on the boundary, hence this time we’ll have a constraint problem. We want the

extrema of f(x, y) = xy subject to g(x, y) = x2 + 2y2 − 1 = 0. We use Lagrange multipliers:

grad f = λ grad g and x2 + 2y2 = 1

which give

y = 2λx

x = 4λy

x2 + 2y2 = 1

From the first two equations we getxy = 8λ2xy.

So x = 0 or y = 0 or 8λ2 = 1.If x = 0, from the last equation 2y2 = 1 so y = ±

√2

2and we get the solutions (0,

√2

2) and (0,−

√2

2).

If y = 0, from the last equation we get x2 = 1 and so the solutions are (1, 0) and (−1, 0).If x 6= 0 and y 6= 0 then 8λ2 = 1, hence λ = ± 1

2√

2. For λ = 1

2√

2

x =√

2y

and plugging into the third equation gives 4y2 = 1 so we get the solutions (√

22, 1

2) and (−

√2

2,− 1

2).

For λ = − 1

2√

2we get

x = −√

2y

and plugging into the third equation gives 4y2 = 1, and the solutions (√

22,− 1

2) and (−

√2

2, 1

2). So finally we have the

solutions: (1, 0), (−1, 0), (√

22, 1

2), (−

√2

2,− 1

2), (√

22,− 1

2), (−

√2

2, 1

2).

Evaluating f at these points gives:

f(0,

√2

2) = f(0,−

√2

2) = f(1, 0) = f(−1, 0) = 0

f(

√2

2,

1

2) = f(−

√2

2,−1

2) =

√2

4

f(

√2

2,−1

2) = f(−

√2

2,

1

2) = −

√2

4.

The region x2 + 2y2 ≤ 1 is closed and bounded, so the maximum and minimum values of f in the region exist. Hencethe maximum value of f is

√2

4and the minimum value of f is −

√2

4.

14. The region x2 + y2 ≤ 1 is the shaded disk of radius 1 centered at the origin (including the circle x2 + y2 = 1) shown inFigure 15.20.

Let’s first compute the critical points of f in the interior of the disk. We have

fx = 3x2 = 0

fy = −2y = 0,

whose solution is x = y = 0. So the only one critical point is (0, 0). As fxx(0, 0) = 0, fyy(0, 0) = −2 and fxy(0, 0) =0,

D = fxx(0, 0) · fyy(0, 0)− (fxy(0, 0))2 = 0

which does not tell us anything about the nature of the critical point (0, 0).But, if we choose x,y very small in absolute value and such that x3 > y2, then f(x, y) > 0. If we choose x,y very

small in absolute value and such that x3 < y2, then f(x, y) < 0. As f(0, 0) = 0, we conclude that (0, 0) is a saddlepoint.

We can get the same conclusion looking at the level curves of f around (0, 0), as shown in Figure 15.21.So, f does not have extrema in the interior of the disk.Now, let’s find the local extrema of f on the circle x2 + y2 = 1. So we want the extrema of f(x, y) = x3 − y2

subject to the constraint g(x, y) = x2 + y2 − 1 = 0. Using Lagrange multipliers we get


15.3 SOLUTIONS 1069

which gives

3x2 = 2λx

−2y = 2λy

x2 + y2 = 1.

From the second equation y = 0 or λ = −1.If y = 0, from the third equation we get x2 = 1, which gives the solutions (1, 0), (−1, 0).If y 6= 0 then λ = −1 and from the first equation we get 3x2 = −2x, hence x = 0 or x = − 2

3. If x = 0, from the

third equation we get y2 = 1, so the solutions (0, 1),(0,−1). If x = − 23

, from the third equation we get y2 = 59

, so thesolutions (− 2

3,√

53

), (− 23,−√

53

).Evaluating f at these points we get

f(1, 0) = 1, f(−1, 0) = f(0, 1) = f(0,−1) = −1

and

f

(−2

3,−√

5

3

)= f

(−2

3,

√5

3

)= −23

27.

The region x2 +y2 ≤ 1 is closed and bounded, so maximum and minimum values of f in the region exist. Thereforethe maximum value of f is 1 and the minimum value is −1.

1

1

x

y

Figure 15.20

−1 1

−1

1

0 0.3 0.7−0.1−0.3

−0.7

x

y

Figure 15.21: Level curves of f

15. The region x+ y ≥ 1 is the shaded half plane (including the line x+ y = 1) shown in Figure 15.22.

1 2 3

1

2

3

13

5

−1−3−5

x

y

Figure 15.22

Let’s look for the critical points of f in the interior of the region. As

fx = 3x2

fy = 1


there are no critical points inside the shaded region. Now let’s find the extrema of f on the boundary of our region. Wewant the extrema of f(x, y) = x3 + y subject to the constraint g(x, y) = x+ y − 1 = 0. We use Lagrange multipliers

grad f = λ grad g and x+ y = 1,

which give

3x2 = λ

1 = λ

x+ y = 1.

From the first two equations we get 3x2 = 1, so the solutions are

(1√3, 1− 1√

3) and (− 1√

3, 1 +

1√3

).

Evaluating f at these points we get

f(1√3, 1− 1√

3) = 1− 2

3√

3

f(− 1√3, 1 +

1√3

) = 1 +2

3√

3.

From the contour diagram in Figure 15.22, we see that ( 1√3, 1 − 1√

3) is a local minimum and (− 1√

3, 1 + 1√

3) is a local

maximum of f on x+ y = 1. Are they global extrema as well?If we take x very big and y = 1− x then f(x, y) = x3 + y = x3 − x+ 1 which can be made as big as we want (if

we choose x big enough). So there will be no global maximum.Similarly, taking x negative with big absolute value and y = 1− x, f(x, y) = x3 + y = x3 − x+ 1 can be made as

small as we want (if we choose x small enough). So there is no global minimum. This can also be seen from Figure 15.22.

Problems

16. We know that a maximum or minimum value of f subject to the constraint equation g(x, y) = c occurs where grad fis parallel to grad g, or at the endpoints of the constraint. The vectors grad f and grad g are parallel where the graph ofg(x, y) = c is tangent to the contours of f , which occurs at approximately x = 6 and y = 6. At the point (6, 6), wehave f = 400. The graph of g(x, y) = c crosses the contours f = 300, f = 200, f = 100 but does not cross anycontours with f -values greater than 400. We see that the maximum of f subject to the constraint is 400 at the point (6, 6).It appears that f takes on its minimum value (less than 100) at one of the endpoints, which are approximately (10.5, 0)and (0, 13.5).

17. (a) The contour for z = 1 is the line 1 = 2x + y, or y = −2x + 1. The contour for z = 3 is the line 3 = 2x + y, ory = −2x+ 3. The contours are all lines with slope −2. See Figure 15.23.

−1−2−3−4−5 1 2 3 4 5−1

−2

−3

−4

−5

1

2

3

4

5

−7−5

−3−1

13

57

x

y

Figure 15.23

−1−2−3−4−5 1 2 3 4 5−1

−2

−3

−4

−5

1

2

3

4

5

−7−5

−3−1

13

57

x

y

Figure 15.24

15.3 SOLUTIONS 1071

(b) The graph of x2 + y2 = 5 is a circle of radius√

5 = 2.236 centered at the origin. See Figure 15.24.(c) The circle representing the constraint equation in Figure 15.24 appears to be tangent to the contour close to z = 5 at

the point (2, 1), and this is the contour with the highest z-value that the circle intersects. The circle is tangent to thecontour z = −5 approximately at the point (−2,−1), and this is the contour with the lowest z-value that the circleintersects. Therefore, subject to the constraint x2 + y2 = 5, the function f has a maximum value of about 5 at thepoint (2, 1) and a minimum value of about −5 at the point (−2,−1).

Since the radius vector, 2~i + ~j , at the point (2, 1) is perpendicular to the line 2x + y = 5, the maximum isexactly 5 and occurs at (2, 1). Similarly, the minimum is exactly −5 and occurs at (−2,−1).

(d) The objective function is f(x, y) = 2x+ y and the constraint equation is g(x, y) = x2 + y2 = 5, and so grad f =2~i +~j and grad g = (2x)~i + (2y)~j . Setting grad f = λ grad g gives

2 = λ(2x),

1 = λ(2y).

On the constraint, x 6= 0 and y 6= 0. Thus, from the first equation, we have λ = 1/x, and from the second equationwe have λ = 1/(2y). Setting these equal gives

x = 2y.

Substituting this into the constraint equation x2 +y2 = 5 gives (2y)2 +y2 = 5 so y = −1 and y = 1. Since x = 2y,the maximum or minimum values occur at (2, 1) or (−2,−1). Since f(2, 1) = 5 and f(−2,−1) = −5, the functionf(x, y) = 2x+ y subject to the constraint x2 + y2 = 5 has a maximum value of 5 at the point (2, 1) and a minimumvalue of −5 at the point (−2,−1). This confirms algebraically what we observed graphically in part (c).

18. We want to minimizeC = f(q1, q2) = 2q2

1 + q1q2 + q22 + 500

subject to the constraint q1 + q2 = 200 or g(q1, q2) = q1 + q2 − 200 = 0.Since∇f = (4q1 + q2)~i + (2q2 + q1)~j and∇g =~i +~j ,∇f = λ∇g gives

4q1 + q2 = λ

2q2 + q1 = λ.

Solving we get4q1 + q2 = 2q2 + q1

so3q1 = q2.

We wantq1 + q2 = 200

q1 + 3q1 = 4q1 = 200.

Thereforeq1 = 50 units, q2 = 150 units.

19.r

6

?

h

-

Figure 15.25

Let V be the volume and S be the surface area of the container. Then

V = πr2h and S = 2πrh+ 2πr2


where h is the height and r is the radius as shown in Figure 15.25. We have V = 100 cm3 as our constraint. Since

∇S = (2πh+ 4πr)~i + 2πr~j = π((2h+ 4r)~i + 2r~j )

and ∇V = 2πrh~i + πr2~j = π(2rh~i + r2~j ),

at the optimum

∇S = λ∇V, we have

π((2h+ 4r)~i + 2r~j ) = πλ(2rh~i + r2~j ),

that is 2h+ 4r = 2λrh and 2r = λr2, hence λ =2

r.

We assume r 6= 0 or else we have a very awkward cylinder. Then, plug λ = 2/r into the first equation to obtain:

2h+ 4r = 2(

2

r

)rh

2h+ 4r = 4h

h = 2r.

Finally, solve for r and h using the constraint:

V = πr2h = 100

πr2(2r) = 100

r3 =50

π

r =3

√50

π.

Solving for h, we obtain h = 2r = 23

√50

π.

20. Constraint is G = P1x+ P2y −K = 0.Since∇Q = λ∇G, we have

caxa−1yb = λP1 and cbxayb−1 = λP2.

Dividing the two equations yieldscaxa−1yb

cbxayb−1=λP1

λP2, or simplifying,

ay

bx=P1

P2. Hence, y =

bP1

aP2x.

Substitute into the constraint to obtain P1x+ P2bP1

aP2x = P1

(a+ b

a

)x = K, giving

x =aK

(a+ b)P1and y =

bK

(a+ b)P2.

We now check that this is indeed the maximization point. Since x, y ≥ 0, possible maximization points are (0,K

P2),

(K

P1, 0), and (

aK

(a+ b)P1,

bK

(a+ b)P2). Since Q = 0 for the first two points and Q is positive for the last point, it follows

that (aK

(a+ b)P1,

bK

(a+ b)P2) gives the maximal value.

21. (a) The company wishes to maximize P (x, y) given the constraint C(x, y) = 50, 000. The objective function is P (x, y)and the constraint equation is C(x, y) = 50, 000. The Lagrange multiplier λ is approximately equal to the change inP (x, y) given a one unit increase in the budget constraint. In other words, if we increase the budget by $1, we canproduce about λ more units of the good.

(b) The company wishes to minimize C(x, y) given the constraint equation P (x, y) = 2000. The objective function isC(x, y) and the constraint equation is P (x, y) = 2000. The Lagrange multiplier λ is approximately equal to thechange in C(x, y) given a one unit increase in the production constraint. In other words, it costs about λ dollars toproduce one more unit of the good.

15.3 SOLUTIONS 1073

22. The company wants to maximize f(x, y) = 500x0.6y0.3 given the constraint g(x, y) = 10x + 25y = 2000. Settinggrad f = λ grad g gives

500(0.6x−0.4)y0.3 = 10λ,

500x0.6(0.3y−0.7) = 25λ.

From the first equation we have λ = 30y0.3/x0.4, and from the second equation we have λ = 6x0.6/y0.7. Setting theseequal gives

y = 0.2x.

Substituting this into the constraint equation 10x+ 25y = 2000 gives x = 133.33. Since y = 0.2x, the maximum valueoccurs at x = 133.33 and y = 26.67.

(a) The company should purchase 133.33 units of chemical X and 26.67 units of chemical Y. With these purchases, thecompany will be able to produce f(133.33, 26.67) = 25, 219 units of chemical Z.

(b) When x = 133.33 and y = 26.67, we see that λ = 11.348. If $1 is added to the budget, the company will be able toproduce about 11.348 additional units of chemical Z.

23. (a) Let c be the cost of producing the product. Then c = 10W + 20K = 3000. At optimum production,

∇q = λ∇c.

∇q =(

92W−

14K

14

)~i +

(32W

34K−

34

)~j , and∇c = 10~i + 20~j . Equating we get

92W−

14K

14 = λ10, and 3

2W

34K−

34 = λ20.

Dividing yields K = 16W , so substituting into c gives

10W + 20(

1

6W)

=40

3W = 3000.

Thus W = 225 and K = 37.5. Substituting both answers to find λ gives

λ =92(225)−

14 (37.5)

14

10= 0.2875.

We also find the optimum quantity produced, q = 6(225)34 (37.5)

14 = 862.57.

(b) At the optimum values found above, marginal productivity of labor is given by

∂q

∂W

∣∣∣∣(225,37.5)

=9

2W−

14K

14

∣∣∣∣(225,37.5)

= 2.875,

and marginal productivity of capital is given by

∂q

∂K

∣∣∣∣(225,37.5)

=3

2W

34K−

34

∣∣∣∣(225,37.5)

= 5.750.

The ratio of marginal productivity of labor to that of capital is

∂q∂W∂q∂K

=1

2=

10

20=

cost of a unit of Lcost of a unit of K

.

(c) When the budget is increased by one dollar, we substitute the relationK1 = 16W1 into 10W1 +20K1 = 3001 which

gives 10W1 + 20( 16W1) = 40

3W1 = 3001. Solving yields W1 = 225.075 and K1 = 37.513, so q1 = 862.86 =

q + 0.29. Thus production has increased by 0.29 ≈ λ, the Lagrange multiplier.

24. (a) The problem is to maximizeV = 1000D0.6N0.3

subject to the budget constraint in dollars

40000D + 10000N ≤ 600000

or (in thousand dollars)40D + 10N ≤ 600


(b) Let B = 40D + 10N = 600 (thousand dollars) be the budget constraint. At the optimum

∇V = λ∇B,

so∂V

∂D= λ

∂B

∂D= 40λ

∂V

∂N= λ

∂B

∂N= 10λ.

Thus∂V∂D∂V∂N

= 4.

Therefore, at the optimum point, the rate of increase in the number of visits to the number of doctors is four times thecorresponding rate for nurses. This factor of four is the same as the ratio of the salaries.

(c) Differentiating and setting∇V = λ∇B yields

600D−0.4N0.3 = 40λ

300D0.6N−0.7 = 10λ

Thus, we get600D−0.4N0.3

40= λ =

300D0.6N−0.7

10So

N = 2D.

To solve for D and N , substitute in the budget constraint:

600− 40D − 10N = 0

600− 40D − 10 · (2D) = 0

So D = 10 and N = 20.

λ =600(10−0.4)(200.3)

40≈ 14.67

Thus the clinic should hire 10 doctors and 20 nurses. With that staff, the clinic can provide

V = 1000(100.6)(200.3) ≈ 9,779 visits per year.

(d) From part c), the Lagrange multiplier is λ = 14.67. At the optimum, the Lagrange multiplier tells us that about 14.67extra visits can be generated through an increase of $1,000 in the budget. (If we had written out the constraint indollars instead of thousands of dollars, the Lagrange multiplier would tell us the number of extra visits per dollar.)

(e) The marginal cost, MC, is the cost of an additional visit. Thus, at the optimum point, we need the reciprocal of theLagrange multiplier:

MC =1

λ≈ 1

14.67≈ 0.068 (thousand dollars)

i.e. at the optimum point, an extra visit costs the clinic 0.068 thousand dollars, or $68.This production function exhibits declining returns to scale (e.g. doubling both inputs less than doubles output,

because the two exponents add up to less than one). This means that for large V , increasing V will require increasingD and N by more than when V is small. Thus the cost of an additional visit is greater for large V than for small. Inother words, the marginal cost will rise with the number of visits.

25. (a) The solution to Problem 23 gives λ = 0.29. We recalculate λ with a budget of $4000.The condition that grad q = λ grad(budget) in Problem 23 gives

9

2W−1/4K1/4 = λ(10) and

3

2W 3/4K−3/4 = λ(20),

so K = 16W . Substituting into the budget constraint after replacing the budget of $3000 by $4000 gives

10W + 20(1

6W ) =

40

3W = 4000.

Thus, W = 300 and K = 50 and q = 1150.098.Multiplying the first equation by W and the second by K and adding gives

W (9

2W−1/4K1/4) +K(

3

2W 3/4K−3/4) = W (10λ) +K(20λ).

15.3 SOLUTIONS 1075

So(

9

2+

3

2

)W 3/4K1/4 = λ(10W + 20K)

6W 3/4K1/4 = λ(4000)

Thus,

λ =6W 3/4K1/4

4000=

1150.098

4000= 0.29

Thus, the value of λ remains unchanged.(b) The solution to Problem 24 shows that λ = 14.67. We solve the problem again with a budget of $700,000.

The condition that gradV = λ gradB in Problem 24 gives

600D−0.4N0.3 = 40λ

300D0.6N−0.7 = 10λ

Thus,N = 2D. Substituting in the budget constraint after replacing the budget of 600 by 700 (the budget in measuredin thousands of dollars) gives

40D + 10(2D) = 700

so D = 11.667 and N = 23.337 and V = 11234.705. As in part a), we multiply the first equation by D and thesecond by N and add:

D(600D−0.4N0.3) +N(300D0.6N−0.7) = D(40λ) +N(10λ),

so

(600 + 300)D0.6N0.3 = λ(400 + 10N)

900D0.6N0.3 = λ(700)

Since V = 1000D0.6N0.3 = 11234.705, we have

λ =900D0.6N0.3

700=

9

7(V

1000) = 14.44.

Thus, the value of λ has changed with the budget.(c) We are interested in the marginal increase of production with budget (that is, the value of λ) and whether it is affected

by the budget.Suppose $B is the budget. In part (a) we found

λ =6W 3/4K1/4

B

and in part (b) we found

λ =900D0.6N0.3

B.

In part (a), both W and K are proportional to B. Thus, W = c1B and K = c2B, so

λ =6(c1B)3/4(c2B)1/4

B

=6c

3/41 C

1/42 B3/4B1/4

B

= 6c3/41 c

1/42 .

So we see λ is independent of B.In part (b), both D and N are proportional to B, so D = c3B and N = c4B. Thus,

λ =900(c3B)0.6(c4B)0.3

B

=900c0.63 C0.3

4 B0.6B0.3

B

= 900c0.63 c0.341

B0.1.


So we see λ is not independent of B.The crucial difference is that the exponents in Problem 23 add to 1, that is 3/4+1/4 = 1, whereas the exponents

in Problem 24 do not add to 1, since 0.6 + 0.3 = 0.9.Thus, the condition that must be satisfied by the Cobb-Douglas production function

Q = cKaLb

to ensure that the value of λ is not affected by production is that

a+ b = 1.

This is called constant returns to scale.

26. (a) The curves are shown in Figure 15.26.

20 40 60 80 100

500

1000

1500

III

III

(50, 500)

s = 1000− 10l

l

s

Figure 15.26

(b) The income equals $10/hour times the number of hours of work:

s = 10(100− l) = 1000− 10l.

(c) The graph of this constraint is the straight line in Figure 15.26.(d) For any given salary, curve III allows for the most leisure time, curve I the least. Similarly, for any amount of leisure

time, curve III also has the greatest salary, and curve I the least. Thus, any point on curve III is preferable to any pointon curve II, which is preferable to any point on curve I. We prefer to be on the outermost curve that our constraintallows. We want to choose the point on s = 1000 − 10l which is on the most preferable curve. Since all the curvesare concave up, this occurs at the point where s = 1000− 10l is tangent to curve II. So we choose l = 50, s = 500,and work 50 hours a week.

27. The maximum of f(x, y) = ax2 + bxy + cy2 subject to the constraint g(x, y) = 1 where g(x, y) = x2 + y2 occurswhere grad f = λ grad g. Since grad f = (2ax+ by)~i + (bx+ 2cy)~j and grad g = 2x~i + 2y~j we have

2ax+ by = 2xλ

bx+ 2cy = 2yλ

x2 + y2 = 1

Adding x times the first equation to y times the second gives x(2ax+ by) + y(bx+ 2cy) = (2x2 + 2y2)λ. Dividing by2 and using the constraint equation gives f(x, y) = ax2 + bxy + cy2 = (x2 + y2)λ = λ. This equation holds for allsolutions (x, y, λ) of the three equations, including the solution that corresponds to the maximum value of f subject tothe constraint. Thus the maximum value is f(x, y) = λ.

28. (a) Let f(x1, x2, x3) =∑3

i=1xi

2 = x12 + x2

2 + x32 and g(x1, x2, x3) =

∑3

i=1xi = 1. Then grad f = λ grad g

gives2x1 = λ and 2x2 = λ and 2x3 = λ.

sox1 = x2 = x3 =

λ

2.

15.3 SOLUTIONS 1077

Then x1 + x2 + x3 = 1 gives

3λ

2= 1 so λ =

2

3so x1 = x2 = x3 =

1

3.

These values of x1, x2, x3 give the minimum (rather than maximum) because the value of f increases without boundas x2, x2, x3 →∞.

(b) A similar argument shows that∑n

i=1xi has its minimum value subject to

∑n

i=1xi = 1 when

x1 = x2 = · · · = xn =1

n.

29. (a) The gradient vectors,∇f , point inward around a local maximum. See the two points marked A in Figure 15.27.(b) Some of the gradient vectors around a saddle are pointing inward toward the point; some are pointing outward away

from the point. See the point marked B in Figure 15.27.(c) The critical points on g = 1 are at points where ∇f is perpendicular to the curve g = 1. There are four of them, all

marked with a dot in Figure 15.27. Imagine the level surfaces of f sketched in everywhere perpendicular to ∇f ; themaximum value of f is at the point marked C in Figure 15.27

(d) Again imagine level curves of f . The minimum value of f is at the point marked D.

A A

C

B

D

Figure 15.27

(e) At C, the maximum on g = 1, the vector ∇g points outward (because it points toward g = 2), while ∇f pointsinward. The Lagrange multiplier, λ, is defined so that∇f = λ∇g, so λ must be negative.

30. We want to minimize the function h(x, y) subject to the constraint that

g(x, y) = x2 + y2 = 1,0002 = 1,000,000.

Using the method of Lagrange multipliers, we obtain the following system of equations:

hx = −10x+ 4y

10,000= 2λx,

hy = −4x+ 4y

10,000= 2λy,

x2 + y2 = 1,000,000.

Multiplying the first equation by y and the second by x we get

−y(10x+ 4y)

10,000=−x(4x+ 4y)

10,000.


Hence:2y2 + 3xy − 2x2 = (2y − x)(y + 2x) = 0,

and so the climber either moves along the line x = 2y or y = −2x.We must now choose one of these lines and the direction along that line which will lead to the point of minimum

height on the circle. To do this we find the points of intersection of these lines with the circle x2 + y2 = 1,000,000,compute the corresponding heights, and then select the minimum point.

If x = 2y, the third equation gives5y2 = 1,0002,

so that y = ±1,000/√

5 ≈ ±447.21 and x = ±894.43. The corresponding height is h(±894.43,±447.21) = 2400 m.If y = −2x, we find that x = ±447.21 and y = ∓894.43. The corresponding height is h(±447.21,∓894.43) =2900 m. Therefore, she should travel along the line x = 2y, in either of the two possible directions.

31. The objective function isf(x, y, z) =

√(x− a)2 + (y − b)2 + (z − c)2,

and the constraint isg(x, y, z) = Ax+By + Cz +D = 0.

Partial derivatives of f and g are

fx =12· 2 · (x− a)

f(x, y, z)=

x− af(x, y, z)

,

fy =12· 2 · (y − b)f(x, y, z)

=y − b

f(x, y, z),

fz =12· 2 · (z − c)f(x, y, z)

=z − c

f(x, y, z),

gx = A, gy = B, and gz = C.

Using Lagrange multipliers, we need to solve the equations

grad f = λ grad g

where grad f = fx~i + fy~j + fz~k and grad g = gx~i + gy~j + gz~k . This gives a system of equations:

x− af(x, y, z)

= λA

y − bf(x, y, z)

= λB

z − cf(x, y, z)

= λC

Ax+By + Cz +D = 0.

Now x−aA

= y−bB

= z−cC

= λf(x, y, z) gives

x =A

B(y − b) + a,

z =C

B(y − b) + c,

Substitute into the constraint,

A(A

B(y − b) + a

)+By + C

(C

B(y − b) + c

)+D = 0,

(A2

B+B +

C2

B

)y =

A2

Bb−Aa+

C2

Bb− Cc−D.

Hence

15.3 SOLUTIONS 1079

y =(A2 + C2)b−B(Aa+ Cc+D)

A2 +B2 + C2,

y − b =−B(Aa+Bb+ Cc+D)

A2 +B2 + C2

x− a =A

B(y − b)

=−A(Aa+Bb+ Cc+D)

A2 +B2 + C2

z − c =C

B(y − b)

=−C(Aa+Bb+ Cc+D)

A2 +B2 + C2

Thus the minimum f(x, y, z) is

f(x, y, z) =√

(x− a)2 + (y − b)2 + (z − c)2

=[(−A(Aa+Bb+ Cc+D)

A2 +B2 + C2

)2

+

(−B(Aa+Bb+ Cc+D)

A2 +B2 + C2

)2

+

(−C(Aa+Bb+ Cc+D)

A2 +B2 + C2

)2 ]1/2

=|Aa+Bb+ Cc+D|√

A2 +B2 + C2.

The geometric meaning is finding the shortest distance from a point (a, b, c) to the plane Ax+By + Cz +D = 0.

32. (a) The objective function is the energy loss, i21R1 + i22R2, and the constraint is i1 + i2 = I , where I is a constant. TheLagrangian function is

L(i1, i2, λ) = i21R1 + i22R2 − λ(i1 + i2 − I).

We look for solutions to the system of equations we get from gradL = ~0 :

∂L∂i1

= 2i1R1 − λ = 0

∂L∂i2

= 2i2R2 − λ = 0

∂L∂λ

= −(i1 + i2 − I) = 0.

Combining∂L∂i1− ∂L∂i2

= 2(i1R1 − i2R2) = 0 with∂L∂λ

= 0 gives the two equation system

i1R1 − i2R2 = 0

i1 + i2 = I.

Substituting i2 = I − i1 into the first equation leads to

i1 =R2

R1 +R2I

i2 =R1

R1 +R2I.

(b) Ohm’s Law states that across a resistor

Voltage = Current · Resistance.

Since λ/2 = i1 ·R1 = i2 ·R2, the Lagrange multiplier λ equals twice the voltage across the resistors.


33. (a) We draw the level curves (parallel straight lines) of f(x, y) = ax + by + c. We can see that the level lines with themaximum and minimum f -values which intersect with the disk are the level lines that are tangent to the boundary ofthe disk. Therefore, the maximum and minimum occur at the boundary of the disk. See Figure 15.28.

f = max

f = min

f increases

M

Figure 15.28f = max

f = min

f increases]

Figure 15.29

f = max

f = minf increases 6

Figure 15.30

(b) Similar to part (a), we see the level lines with the largest and smallest f -values which intersect with the rectangle mustpass the corner of the rectangle. So the maximum and minimum occur at the corners of rectangle. See Figure 15.29.When the level curves are parallel to a pair of the sides, then the points on the sides are all maximum or minimum, asshown below in Figure 15.30.

(c) The graph of f is a plane. The part of the graph lying above a disk R is either a flat disk, in which case every point isa maximum, or is a tilted ellipse, in which case you can see that the maximum will be on the edge. Similarly, the partlying above a rectangle is either a rectangle or a tilted parallelogram, in which case the maximum will be at a corner.

34. The point P is the solution to the constraint optimization problem of maximizing the square of the distance function.

D = x2 + y2 + x2

subject to the constraintg(x, y, z) = f(x, y)− z = 0.

(We take the square of the distance between the point (x, y, z) and the origin, which is

Distance =√

(x− 0)2 + (y − 0)2 + (z − 0)2 =√x2 + y2 + z2,

because it makes the calculations easier.) Therefore, at point P , we have∇D = λ∇g, so∇D is parallel to∇g.We know that ∇g is perpendicular to the surface g(x, y, z) = 0; that is, perpendicular to the surface z = f(x, y).

Also∇D = 2x~i + 2y~j + 2z~k .

15.3 SOLUTIONS 1081

At point P , whose position vector is ~p = a~i + b~j + c~k , we have

∇D = 2(a~i + b~j + c~k ) = 2~p .

Thus, ~p is parallel to∇D and therefore ~p is also perpendicular to the surface.

35. (a) The objective function f(x, y) = px+ qy gives the cost to buy x units of input 1 at unit price p and y units of input2 at unit price q.

The constraint g(x, y) = u tells us that we are only considering the cost of inputs x and y that can be used toproduce quantity u of the product.

Thus the number C(p, q, u) gives the minimum cost to the company of producing quantity u if the inputs itneeds have unit prices p and q.

(b) The Lagrangian function isL(x, y, λ) = px+ qy − λ(xy − u).


∂L∂x

= p− λy = 0

∂L∂y

= q − λx = 0

∂L∂λ

= −(xy − u) = 0.

We see that λ = p/y = q/x so y = px/q. Substituting for y in the constraint xy = u leads to x =√qu/p,

y =√pu/q and λ =

√pq/u. The minimum cost is thus

C(p, q, u) = p

√qu

p+ q

√pu

q= 2√pqu.

36. (a) The objective function U(x, y) gives the utility to the consumer of x units of item 1 and y units of item 2.Since px + qy gives the cost to buy x units of item 1 at unit price p and y units of item 2 at unit price q, the

constraint px+ qy = I tells us that we are only considering the utility of inputs x and y that can be purchased withbudget I .

Thus the number V (p, q, I) gives the maximum utility the consumer can get with a budget of I if the two itemshave unit prices p and q.

The indirect utility function tells how much utility the consumer can buy, depending on his budget and the pricesof the two items.

(b) The value of the Lagrange multiplier λ is the rate of change of the maximum utility V the consumer can get withhis budget as the budget increases. This means that for small changes ∆I in the budget, smart buying will result in achange ∆V ≈ λ∆I in the utility to the consumer of his purchases.

(c) The Lagrangian function isL(x, y, λ) = xy − λ(px+ qy − I).


∂L∂x

= y − λp = 0

∂L∂y

= x− λq = 0

∂L∂λ

= −(px+ qy − I) = 0.

We see that λ = y/p = x/q so y = px/q. Substituting for y in the constraint px + qy = I leads to x = I/(2p),y = I/(2q) and λ = I/(2pq). The maximum utility is thus

V (p, q, I) = U(I

2p,I

2q) =

I

2p· I

2q=

I2

4pq.

The marginal utility of money is

λ(p, q, I) = λ =I

2pq.


37. (a) The critical points of h(x, y) occur where

hx(x, y) = 2x− 2λ = 0

hy(x, y) = 2y − 4λ = 0.

The only critical point is (x, y) = (λ, 2λ) and it gives a minimum value for h(x, y). That minimum value ism(λ) =h(λ, 2λ) = λ2 + (2λ)2 − λ(2λ+ 4(2λ)− 15) = −5λ2 + 15λ.

(b) The maximum value of m(λ) = −5λ2 + 15λ occurs at a critical point, where m′(λ) = −10λ + 15 = 0. At thispoint, λ = 1.5 and m(λ) = −5 · 1.52 + 15 · 1.5 = 11.25.

(c) We want to minimize f(x, y) = x2 + y2 subject to the constraint g(x, y) = 15, where g(x, y) = 2x + 4y. TheLagrangian function is L(x, y, λ) = x2 + y2 − λ(2x+ 4y − 15) so we solve the system of equations

∂L∂x

= 2x− 2λ = 0

∂L∂y

= 2y − 4λ = 0

∂L∂λ

= −(2x+ 4y − 15) = 0.

The first two equations give x = λ and y = 2λ. Substitution into the third equation gives 2λ + 4(2λ) − 15 = 0or λ = 1.5. Thus x = 1.5 and y = 3. The minimum value of f(x, y) subject to the constraint is f(1.5, 3) =1.53 + 32 = 11.25.

(d) The two question have the same answer.

38. You should try to anticipate your opponent’s choice. After you choose a value λ, your opponent will use calculus tofind the point (x, y) that maximizes the function f(x, y) = 10 − x2 − y2 − 2x − λ(2x + 2y). At that point, we havefx = −2x− 2− 2λ = 0 and fy = −2y− 2λ = 0, so your opponent will choose x = −1−λ and y = −λ. This gives avalue L(−1−λ,−λ, λ) = 10−(−1−λ)2−(−λ)2−2(−1−λ)−λ(2(−1−λ)+2(−λ)) = 11+2λ+2λ2 which youwant to make as small as possible. You should choose λ to minimize the function h(λ) = 11 + 2λ+ 2λ2. You choose λso that h′(λ) = 2 + 4λ = 0, or λ = −1/2. Your opponent then chooses (x, y) = (−1− λ,−λ) = (−1/2, 1/2), givinga final score of L(−1/2, 1/2,−1/2) = 10.5. No choice of λ that you can make can force the value of L below 10.5. Butyour choice of λ = −1/2 makes it impossible for your opponent to force the value of L above 10.5.

Solutions for Chapter 15 Review

Exercises

1. The critical points of f are obtained by solving fx = fy = 0, that is

fx(x, y) = 2y2 − 2x = 0 and fy(x, y) = 4xy − 4y = 0,

so2(y2 − x) = 0 and 4y(x− 1) = 0

The second equation gives either y = 0 or x = 1. If y = 0 then x = 0 by the first equation, so (0, 0) is a critical point. Ifx = 1 then y2 = 1 from which y = 1 or y = −1, so two further critical points are (1,−1), and (1, 1).

SinceD = fxxfyy − (fxy)2 = (−2)(4x− 4)− (4y)2 = 8− 8x− 16y2,

we haveD(0, 0) = 8 > 0, D(1, 1) = D(1,−1) = −16 < 0,

and fxx = −2 < 0. Thus, (0, 0) is a local maximum; (1, 1) and (1,−1) are saddle points.

2. At a critical point

fx(x, y) = 2xy − 2y = 0

fy(x, y) = x2 + 4y − 2x = 0.

SOLUTIONS to Review Problems for Chapter Fifteen 1083

From the first equation, 2y(x− 1) = 0, so either y = 0 or x = 1. If y = 0, then x2 − 2x = 0, so x = 0 or x = 2. Thus(0, 0) and (2, 0) are critical points. If x = 1, then 12 + 4y − 2 = 0, so y = 1/4. Thus (1, 1/4) is a critical point. Now

D = fxxfyy − (fxy)2 = 2y · 4− (2x− 2)2 = 8y − 4(x− 1)2,

soD(0, 0) = −4, D(2, 0) = −4, D(1, 1

4) = 2

so (0, 0) and (2, 0) are saddle points. Since fyy = 4 > 0, we see that (1, 1/4) is a local minimum.

3. At a critical point

fx(x, y) = 6x2 − 6xy + 12x = 0

fy(x, y) = −3x2 − 12y = 0

From the second equation, we conclude that −3(x2 + 4y) = 0, so y = − 14x2. Substituting for y in the first equation

gives

6x2 − 6x(−1

4x2)

+ 12x = 0

orx2 +

1

4x3 + 2x =

x

4(4x+ x2 + 8) = 0.

Thus x = 0 or x2 + 4x+ 8 = 0. The quadratic has no real solutions, so the only one critical point is (0, 0).At (0, 0), we have

D(0, 0) = fxxfyy − (fxy)2 = (12)(−12)− 02 = −144 < 0,

so (0, 0) is a saddle point.


fx = cosx+ cos (x+ y).

fy = cos y + cos (x+ y).

Setting fx = 0 and fy = 0 givescosx = cos y

For 0 < x < π and 0 < y < π, cosx = cos y only if x = y. Then, setting fx = fy = 0:

cosx+ cos 2x = 0,

cosx+ 2 cos2 x− 1 = 0,

(2 cosx− 1)(cosx+ 1) = 0.

So cosx = 1/2 or cosx = −1, that is x = π/3 or x = π. For the given domain 0 < x < π, 0 < y < π, we onlyconsider the solution when x = π/3 then y = x = π/3. Therefore, the critical point is ( π

3, π

3).

Sincefxx(x, y) = − sinx− sin (x+ y) fxx(π

3, π

3) = − sin π

3− sin 2π

3= −√

3

fxy(x, y) = − sin (x+ y) fxy(π3, π

3) = − sin 2π

3= −

√3

2

fyy(x, y) = − sin y − sin (x+ y) fyy(π3, π

3) = − sin π

3− sin 2π

3= −√

3

the discriminant isD(x, y) = fxxfyy − f2

xy

= (−√

3)(−√

3)− (−√

32

)2 = 94> 0.

Since fxx(π3, π

3) = −

√3 < 0, (π

3, π

3) is a local maximum.

5. We find critical points:

fx(x, y) = 12− 6x = 0

fy(x, y) = 6− 2y = 0

so (2, 3) is the only critical point. At this point

D = fxxfyy − (fxy)2 = (−6)(−2) = 12 > 0,

and fxx < 0, so (2, 3) is a local maximum. Since this is a quadratic, the local maximum is a global maximum.Alternatively, we complete the square, giving

f(x, y) = 10− 3(x2 − 4x)− (y2 − 6y) = 31− 3(x− 2)2 − (y − 3)2.

This expression for f shows that its maximum value (which is 31) occurs where x = 2, y = 3.


6. The partial derivatives are fx = 2x − 3y, fy = 3y2 − 3x. For critical points, solve fx = 0 and fy = 0 simultaneously.From 2x− 3y = 0 we get x = 3

2y. Substituting it into 3y2 − 3x = 0, we have that

3y2 − 3(3

2y) = 3y2 − 9

2y = y(3y − 9

2) = 0.

So y = 0 or 3y − 92

= 0, that is, y = 0 or y = 3/2. Therefore the critical points are (0, 0) and ( 94, 3

2).

The contour diagram for f in Figure 15.31 (drawn by a computer), shows that (0, 0) is a saddle point and that ( 94, 3

2) is a

local minimum.

−4 −3 −2 −1 1 2 3 4

−4

−3

−2

−1

1

2

3

4110

90 70

5030

10

−10

−30

−50−70

−90

10

30

0

0

0

x

y

( 94, 3

2)

Figure 15.31: Contour map of f(x, y) = x2 + y3 − 3xy

We can also see that (0, 0) is a saddle point and ( 94, 3

2) is a local minimum analytically. Since fxx = 2, fyy =

6y, fxy = −3, the discriminant is

D(x, y) = fxxfyy − f2xy = 12y − (−3)2 = 12y − 9.

D(0, 0) = −9 < 0, so (0, 0) is a saddle point.D( 9

4, 3

2) = 9 > 0 and fxx = 2 > 0, we know that ( 9

4, 3

2) is a local minimum. The point ( 9

4, 3

2) is not a global minimum

since f( 94, 3

2) = −1.6875, whereas f(0,−2) = −8.


fx = y +1

x, fy = x+ 2y.

For critical points, solve fx = 0 and fy = 0 simultaneously. From fy = x+ 2y = 0 we get that x = −2y. Substitutinginto fx = 0, we have

y +1

x= y − 1

2y=

1

y(y2 − 1

2) = 0

Since 1y6= 0, y2 − 1

2= 0, therefore

y = ± 1√2

= ±√

2

2,

and x = ∓√

2. So the critical points are(−√

2,√

22

)and(√

2,−√

22

). But x must be greater than 0, so

(−√

2,√

22

)is

not in the domain.The contour diagram for f in Figure 15.32 (drawn by computer), shows that

(√2,−

√2

2

)is a saddle point of f(x, y).


4 5 6 7 8

−4

−3

−2

−1

1

2

3

4

x

y

0

?

−10

6

−10

-f(√

2,−√

22

) = − 212

+ ln√

2

−20

010

2030

Figure 15.32: Contour map of f(x, y) = xy + lnx+ y2 − 10

We can also see that(√

2,−√

22

)is a saddle point analytically.

Since fxx = − 1x2 , fyy = 2, fxy = 1, the discriminant is:


= − 2

x2− 1.

D(√

2,−√

22

)= −2 < 0, so

(√2,−

√2

2

)is a saddle point.

8. Note that the x-axis and the y-axis are not in the domain of f . Since x 6= 0 and y 6= 0, by setting fx = 0 and fy = 0 weget

fx = 1− 1

x2= 0 when x = ±1

fy = 1− 4

y2= 0 when y = ±2

So the critical points are (1, 2), (−1, 2), (1,−2), (−1,−2). Since fxx = 2/x3 and fyy = 8/y3 and fxy = 0, thediscriminant is

D(x, y) = fxxfyy − f2xy =

(2

x3

)(8

y3

)− 02 =

16

(xy)3.

Since D < 0 at the points (−1, 2) and (1,−2), these points are saddle points. Since D > 0 at (1, 2) and (−1,−2) andfxx(1, 2) > 0 and fxx(−1,−2) < 0, the point (1, 2) is a local minimum and the point (−1,−2) is a local maximum.No global maximum or minimum, since f(x, y) increases without bound if x and y increase in the first quadrant; f(x, y)decreases without bound if x and y decrease in the third quadrant.

9. The objective function is f(x, y) = 3x− 4y and the constraint equation is g(x, y) = x2 + y2 = 5, so grad f = 3~i − 4~jand grad g = (2x)~i + (2y)~j . Setting grad f = λ grad g gives

3 = λ(2x),

−4 = λ(2y).

From the first equation we have λ = 3/(2x), and from the second equation we have λ = −2/y. Setting these equal gives

x = −3

4y.

Substituting this into the constraint equation x2 + y2 = 5 gives y2 = 16/5, so y = 4/√

5 and y = −4/√

5. Sincex = − 3

4y, there are two points where a maximum or a minimum might occur:

(−3/√

5, 4/√

5) and (3/√

5,−4/√

5).

Since the constraint is closed and bounded, maximum and minimum values of f subject to the constraint exist. Sincef(−3/

√5, 4/√

5) = −5√

5 and f(3/√

5,−4/√

5) = 5√

5, we see that f has a minimum value at (−3/√

5, 4/√

5) anda maximum value at (3/

√5,−4/

√5).


10. The objective function is f(x, y) = x2 + 2y2 and the constraint equation is g(x, y) = 3x + 5y = 200, so grad f =(2x)~i + (4y)~j and grad g = 3~i + 5~j . Setting grad f = λ grad g gives

2x = 3λ,

4y = 5λ.

From the first equation, we have λ = 2x/3, and from the second equation we have λ = 4y/5. Setting these equal gives

x = 1.2y.

Substituting this into the constraint equation 3x+ 5y = 200 gives y = 23.256. Since x = 1.2y, we have x = 27.907. Amaximum or minimum value of f can occur only at (27.907, 23.256).

We have f(27.907, 23.256) = 1860.484. From Figure 15.33, we see that the point (27.907, 23.256) is a minimumvalue of f subject to the given constraint.

10 20 30 40 50

10

20

30

40

50

1000

1860

3000

4000

(27.9, 23.6)

x

3x+ 5y = 200

y

Figure 15.33

11. The objective function is f(x, y) = 2xy and the constraint equation is g(x, y) = 5x + 4y = 100, so grad f =(2y)~i + (2x)~j and grad g = 5~i + 4~j . Setting grad f = λ grad g gives

2y = 5λ,

2x = 4λ.

From the first equation we have λ = 2y/5, and from the second equation we have λ = x/2. Setting these equal gives

y = 1.25x.

Substituting this into the constraint equation 5x+ 4y = 100 gives x = 10 and y = 12.5. A maximum or minimum valuefor f subject to the constraint can occur only at (10, 12.5).

We have f(10, 12.5) = 250. From Figure 15.34, we see that the point (10, 12.5) gives a maximum.

10 20

10

20

30

(10, 12.5)100

250

400

x

5x+ 4y = 100

y

Figure 15.34


12. We will use the Lagrange multipliers with:Objective function: f(x, y) = −3x2 − 2y2 + 20xyConstraint: g(x, y) = x+ y − 100We first find

∇f = (−6x+ 20y)~i + (−4y + 20x)~j

∇g = ~i +~j .

To optimize f , we must solve the equations

∇f = λ∇g(−6x+ 20y)~i + (−4y + 20x)~j = λ(~i +~j ) = λ~i + λ~j

We have a vector equation, so we equate the coordinates:

−6x+ 20y = λ

20x− 4y = λ.

So − 6x+ 20y = 20x− 4y

24y = 26x

y =13

12x

Substituting into the constraint equation x+ y = 100, we obtain:

x+13

12x = 100

25

12x = 100

x = 48.

Consequently, y = 52, and f(48, 52) = 37,600. The point (48, 52) leads to the extreme value of f(x, y), given thatx + y = 100. Note that f has no minimum on the line x + y = 100 since f(x, 100 − x) = −3x2 − 2(100 − x)2 +20x(100− x) = −25x2 + 2400x− 20000 which goes to −∞ as x goes to ±∞. Therefore, the point (48, 52) gives themaximum value for f on the line x+ y = 100.

13. Our objective function is f(x, y, z) = x2 − y2 − 2z and our equation of constraint is g(x, y, z) = x2 + y2 − z = 0.To optimize f(x, y, z) with Lagrange multipliers, we solve ∇f(x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. Thegradients of f and g are

∇f(x, y, z) = 2x~i − 2y~j − 2~k ,

∇g(x, y, z) = 2x~i + 2y~j − ~k .

We get

2x = 2λx

−2y = 2λy

−2 = −λx2 + y2 = z.

The third equation gives λ = 2 and from the first x = 0, from the second y = 0 and from the fourth z = 0. So the onlysolution is (0, 0, 0), and f(0, 0, 0) = 0.

To see what kind of extreme point is (0, 0, 0), let (a, b, c) be a point which satisfies the constraint, i.e. a2 + b2 = c.Then f(a, b, c) = a2 − b2 − 2c = −a2 − 3b2 ≤ 0. The conclusion is that 0 is the maximum value of f and that there isno minimum.

14. We first find the critical points in the disk

∇z = (8x− y)~i + (8y − x)~j

Setting∇z = 0 gives 8x− y = 0 and 8y − x = 0. The only solution is x = y = 0. So (0, 0) is the only critical point inthe disk.


Next we find the extremal values on the boundary using Lagrange multipliers. We have objective function z =4x2 − xy + 4y2 and constraint G = x2 + y2 − 2 = 0.

∇z = (8x− y)~i + (8y − x)~j

∇G = 2x~i + 2y~j

∇z = λ∇G gives8x− y = 2λx

8y − x = 2λy

If λ = 0 we get

8x− y = 0

8y − x = 0

with only solutions x = y = 0, which does not satisfy the constraint: x2 + y2 − 2 = 0. Therefore λ 6= 0 and we get:

2λy(8x− y) = 2λx(8y − x)

andy(8x− y) = x(8y − x).

So x2 = y2, x = ±y.Substitute into G = 0, we get 2x2 − 2 = 0 so x = ±1. The extremal points on the boundary are therefore

(1, 1), (1,−1), (−1, 1), (−1,−1). The region x2 + y2 ≤ 2 is closed and bounded, so minimum values of f in the regionexist. We check the values of z at these points :

z(1, 1) = 7, z(−1,−1) = 7, z(1,−1) = 9, z(−1, 1) = 9, z(0, 0) = 0

Thus (−1, 1) and (1,−1) give the maxima over the closed disk and (0, 0) gives the minimum.

15. The region x2 ≥ y is the shaded region in Figure 15.35 which includes the parabola y = x2.

10

3050

−10

−30−50

70−70

x

y

Figure 15.35

We first want to find the local maxima and minima of f in the interior of our region. So we need to find the extremaof

f(x, y) = x2 − y2, in the region x2 > y.


fx = 2x = 0

fy = −2y = 0.

As (0, 0) does not belong to the region x2 > y, we have no critical points. Now let’s find the local extrema of f onthe boundary of our region, hence this time we have to solve a constraint problem. We want to find the extrema off(x, y) = x2 − y2 subject to g(x, y) = x2 − y = 0. We use Lagrange multipliers:

grad f = λ grad g and x2 = y.


This gives

2x = 2λx

2y = λ

x2 = y.

From the first equation we get x = 0 or λ = 1.If x = 0, from the third equation we get y = 0, so one solution is (0, 0). If x 6= 0, then λ = 1 and from the second

equation we get y = 12

. This gives x2 = 12

so the solutions ( 1√2, 1

2) and (− 1√

2, 1

2).

So f(0, 0) = 0 and f( 1√2, 1

2) = f(− 1√

2, 1

2) = 1

4. From Figure 15.35 showing the level curves of f and the region

x2 ≥ y, we see that (0, 0) is a local minimum of f on x2 = y, but not a global minimum and that ( 1√2, 1

2) and (− 1√

2, 1

2)

are global maxima of f on x2 = y but not global maxima of f on the whole region x2 ≥ y.So there are no global extrema of f in the region x2 ≥ y.

16. Let the line be in the form y = b + mx. Then, when x equals 0, 1, and 2, y equals b, b + m, and b + 2m respectively.The sum of the squares of the vertical distances, which is what we want to minimize, is

f(m, b) = (4− b)2 + (3− (b+m))2 + (1− (b+ 2m))2

To find critical points, set each partial derivative equal to zero.

fm = 0 + 2(3− (b+m))(−1) + 2(1− (b+ 2m))(−2)

= 6b+ 10m− 10

fb = 2(4− b)(−1) + 2(3− (b+m))(−1) + 2(1− (b+ 2m))(−1)

= 6b+ 6m− 16

Setting both partial derivatives equal to zero and dividing by 2, we get a system of equations:

3b+ 5m = 5

3b+ 3m = 8

with solutions m = − 32

and b = 256

. Thus, the line is y = 256− 3

2x.

Problems

17. Since fxx < 0 and D = fxxfyy − f2xy > 0, the point (1, 3) is a maximum. See Figure 15.36.

1

3

x

y

0

−1−4−16−3

2−64−1

20

Figure 15.36

18. We first express the revenue R in terms of the prices p1 and p2:

R(p1, p2) = p1q1 + p2q2

= p1(517− 3.5p1 + 0.8p2) + p2(770− 4.4p2 + 1.4p1)

= 517p1 − 3.5p21 + 770p2 − 4.4p2

2 + 2.2p1p2.


At a local maximum we have gradR = 0, and so:

∂R

∂p1= 517− 7p1 + 2.2p2 = 0,

∂R

∂p2= 770− 8.8p2 + 2.2p1 = 0.

Solving these equations, we find thatp1 = 110 and p2 = 115.

To see whether or not we have a found a local maximum, we compute the second-order partial derivatives:

∂2R

∂p21

= −7,∂2R

∂p22

= −8.8,∂2R

∂p1∂p2= 2.2.

Therefore,

D =∂2R

∂p21

∂2R

∂p22

− ∂2R

∂p1∂p2= (−7)(−8.8)− (2.2)2 = 56.76,

and so we have found a local maximum point. The graph of P (p1, p2) has the shape of an upside down paraboloid. SinceP is quadratic in q1 and q2, (110, 115) is a global maximum point.

19. Using Lagrange multipliers, let G = 2000− 5x− 10y = 0 be the constraint.

∇P =

(1 +

2xy2

2 · 108

)~i +

(2 +

2yx2

2 · 108

)~j =

(1 +

xy2

108

)~i +

(2 +

yx2

108

)~j .

∇G = −5~i − 10~j .Now,∇P = λ∇G, so

1 +xy2

108= −5λ and 2 +

yx2

108= −10λ.

Thus

2 +2xy2

108= 2 +

yx2

108.

Solving, we get 2y = x or x = 0 or y = 0.

400

200

x

y

ConstraintG = 0

Figure 15.37

From G = 0 we have: when x = 0, y = 200, when y = 0, x = 400, and when x = 2y, x = 200, y = 100. So(0,200), (400,0) and (200,100) are the critical points and they include the end points.Substitute into P : P (0, 200) = 400, P (400, 0) = 400, P (200, 100) = 402 so the organization should buy 200 sacks ofrice and 100 sacks of beans.

20. We want to minimize cost C = 100L+ 200K subject to Q = 900L1/2K2/3 = 36000. Using Lagrange multipliers, weget

∇Q =(450L−1/2K2/3

)~i +

(600L1/2K−1/3

)~j .

∇C = 100~i + 200~j

∇C = λ∇Q gives100 = λ450L−1/2K2/3 and 200 = λ600L1/2K−1/3.

Since λ 6= 0 this gives450L−1/2K2/3 = 300L1/2K−1/3.


Solving, we get L = (3/2)K. Substituting into Q = 36,000 gives

900(

3

2K)1/2

K2/3 = 36,000.

Solving yields K =[40 ·

(23

)1/2]6/7

≈ 19.85, so L ≈ 32(19.85) = 29.78. We can thus calculate cost using K = 20

and L = 30 which gives C = $7, 000.

21. We wish to minimize the objective function

C(x, y, z) = 20x+ 10y + 5z

subject to the budget constraintQ(x, y, z) = 20x1/2y1/4z2/5 = 1, 200.

Therefore, we solve the equations gradC = λ gradQ and Q = 1, 200:

20 = 10λx−1/2y1/4z2/5 or λ = 2x1/2y−1/4z−2/5,

10 = 5λx1/2y−3/4z2/5, or λ = 2x−1/2y3/4z−2/5,

5 = 8λx1/2y1/4z−3/5, or λ = 0.625x−1/2y−1/4z3/5,

20x1/2y1/4z2/5 = 1, 200.

The first and second equations imply thatx = y,

while the second and third equations imply that3.2y = z.

Substituting for x and z in the constraint equation gives

20y1/2y1/4(3.2y)2/5 = 1200

y ≈ 23.47,

and sox ≈ 23.47 and z ≈ 75.1.

22. We want to maximize f(x, y) = 80x0.75y0.25 subject to the budget constraint g(x, y) = 6x + 4y = 8000. Settinggrad f = λ grad g gives us

80(0.75x−0.25)y0.25 = 6λ,

80x0.75(0.25y−0.75) = 4λ.

At the maximum, x, y 6= 0. From the first equation we have λ = 10y0.25/x0.25, and from the second equation we haveλ = 5x0.75/y0.75. Setting these equal gives

y = 0.5x.

Substituting this into the constraint equation 6x+4y = 8000, we see that x = 1000. Since y = 0.5x, we obtain y = 500.That the point (1000, 500) gives the maximum production is suggested by Figure 15.38, since the values of f decrease aswe move along the constraint away from (1000, 500). To produce the maximum quantity, the company should use 1000units of labor and 500 units of capital.


500 1000 1500

500

1000

1500

2000

Budget constraint: 6x+ 4y = 8000

P = (1000, 500)

50000

f = 60000

f = 70000

f = 80000

f = 90000

x

y

Figure 15.38

23. (a) We want to minimize C subject to g = x+ y = 39. Solving∇C = λ∇g gives

10x+ 2y = λ

2x+ 6y = λ

so y = 2x. Solving with x+ y = 39 gives x = 13, y = 26, λ = 182. Therefore C = $4349.(b) Since λ = 182, increasing production by 1 will cause costs to increase by approximately $182. (because λ =‖∇C‖‖∇ g‖ = rate of change of C with g). Similarly, decreasing production by 1 will save approximately $182.

24. (a) To be producing the maximum quantity Q under the cost constraint given, the firm should be using K and L valuesgiven by

∂Q

∂K= 0.6aK−0.4L0.4 = 20λ

∂Q

∂L= 0.4aK0.6L−0.6 = 10λ

20K + 10L = 150.

Hence0.6aK−0.4L0.4

0.4aK0.6L−0.6= 1.5

L

K=

20λ

10λ= 2, so L =

4

3K. Substituting in 20K + 10L = 150, we obtain

20K + 10(

4

3

)K = 150. Then K =

9

2and L = 6, so capital should be reduced by

1

2unit, and labor should be

increased by 1 unit.

(b)New productionOld production

=a4.50.660.4

a50.650.4≈ 1.01, so tell the board of directors, “Reducing the quantity of capital by 1/2 unit

and increasing the quantity of labor by 1 unit will increase production by 1% while holding costs to $150.”

25. Cost of production, C, is given by C = p1W + p2K = b. At the optimal point,∇q = λ∇C.Since∇q =

(c(1− a)W−aKa

)~i +

(caW 1−aKa−1

)~j and∇C = p1

~i + p2~j , we get

c(1− a)W−aKa = λp1 and caW 1−aKa−1 = λp2.

Now, marginal productivity of labor is given by ∂q∂W

= c(1− a)W−aKa and marginal productivity of capital is given by∂q∂K

= caW 1−aKa−1, so their ratio is given by

∂q∂W∂q∂K

=c(1− a)W−aKa

caW 1−aKa−1=λp1

λp2=p1

p2

which is the ratio of the cost of one unit of labor to the cost of one unit of capital.


26. (a) Points A,B,C,D,E; that is, where a level curve of f and the constraint curve are parallel.(b) Point F since the value of f is greatest at this point.(c) Point D has the greatest f value of the points A,B,C,D,E.

9

10

11

12

13

R

g = c

A

B

C

D

E

F

27. (a) By the method of Lagrange multipliers, the point (2, 1) is a candidate when the gradient for f at (2, 1) is a multiple ofthe gradient of the constraint function at (2, 1). The constraint function is g(x, y) = x2+y2, so grad g = 2x~i +2y~j .We have grad g(2, 1) = 4~i + 2~j . This is not a multiple of grad f(2, 1) = −3~i + 4~j , so (2, 1) is not a candidate.

(b) The constraint function is g(x, y) = (x−5)2+(y+3)2, so grad g = 2(x−5)~i +2(y+3)~j . We have grad g(2, 1) =−6~i + 8~j . This is a multiple of grad f(2, 1) = −3~i + 4~j , so (2, 1) is a candidate.

The contours near (2, 1) are parallel straight lines with increasing f -values as we move in the direction of−3~i + 4~j (approximately toward the northwest). The center of the constraint circle is at (5,−3), approximatelysoutheast of the point (2, 1). Thus the point (2, 1) is a candidate for a maximum.

In general, if the constraint is a circle and grad f points outside the circle, then the point is a candidate for amaximum.

(c) The constraint function is g(x, y) = (x+ 1)2 + (y − 5)2. Thus grad g(2, 1) = 6~i − 8~j , which is again a multipleof grad f , so (2, 1) is a candidate.

This time the center of the constraint circle, (−1, 5) is approximately northwest of (2, 1), the same generaldirection in which grad f is pointing. This means the point (2, 1) is a candidate for a minimum.

In general, if the constraint is a circle and grad f points inside the circle, then the point is a candidate for aminimum.

28. (a) (i) Suppose N = kAp. Then the rule of thumb tells us that if A is multiplied by 10, the value of N doubles. Thus

2N = k(10A)p = k10pAp.

Thus, dividing by N = kAp, we have2 = 10p

so taking logs to base 10 we havep = log 2 = 0.3010.

(where log 2 means log10 2). Thus,N = kA0.3010.

(ii) Taking natural logs gives

lnN = ln(kAp)

lnN = ln k + p lnA

lnN ≈ ln k + 0.301 lnA

Thus, lnN is a linear function of lnA.


(b) Table 15.2 contains the natural logarithms of the data:

Table 15.2 lnN and lnA

Island lnA lnN

Redonda 1.1 1.6

Saba 3.0 2.2

Montserrat 2.3 2.7

Puerto Rico 9.1 4.3

Jamaica 9.3 4.2

Hispaniola 11.2 4.8

Cuba 11.6 4.8

Using a least squares fit we find the line:

lnN = 1.20 + 0.32 lnA

This yields the power function:N = e1.20A0.32 = 3.32A0.32

Since 0.32 is pretty close to log 2 ≈ 0.301, the answer does agree with the biological rule.

29. (a) The objective function is the complementary energy,f2

1

2k1+f2

2

2k2, and the constraint is f1 +f2 = mg. The Lagrangian

function is

L(f1, f2, λ) =f2

1

2k1+

f22

2k2− λ(f1 + f2 −mg).


∂L∂f1

=f1

k1− λ = 0

∂L∂f2

=f2

k2− λ = 0

∂L∂λ

= −(f1 + f2 −mg) = 0.

Combining∂L∂f1− ∂L∂f2

=f1

k1− f2

k2= 0 with

∂L∂λ

= 0 gives the two equation system

f1

k1− f2

k2= 0

f1 + f2 = mg.

Substituting f2 = mg − f1 into the first equation leads to

f1 =k1

k1 + k2mg

f2 =k2

k1 + k2mg.

(b) Hooke’s Law states that for a spring

Force of spring = Spring constant · Distance stretched or compressed from equilibrium.

Since f1 = k1 · λ and f2 = k2 · λ, the Lagrange multiplier λ equals the distance the mass stretches the top springand compresses the lower spring.


30. The distance to the origin d =√x2 + y2 + z2. Minimizing d2 minimizes d, so let d2 = f(x, y, z) = x2 + y2 + z2. We

minimize f with constraint z = xy + 1, so we let g(x, y, z) = xy + 1− z.

Lagrange: −→∇f = λ−→∇g gives

2x = λy

2y = λx

2z = −λz = xy + 1

which gives

x = 0

y = 0

z = 1

λ = −2.

The point is (0, 0, 1). This must be a minimum since there are points on z = xy + 1 infinitely far from the origin.(There is no maximum.)

31. Since patient 1 has a visit every x1 weeks, this patient has 1/x1 visits per week. Similarly, patient 2 has 1/x2 visits perweek. Thus, the constraint is

g(x1, x2) =1

x1+

1

x2= m

To minimizef(x1, x2) =

v1

v1 + v2· x1

2+

v2

v1 + v2· x2

2

subject to g(x1, x2), we solve the equations

grad f = λ grad g

g(x1, x2) = m.

This gives us the equations

∂f

∂x1=

v1

v1 + v2· 1

2= λ

(− 1

x21

)= λ

∂g

∂x1

∂f

∂x2=

v2

v1 + v2· 1

2= λ

(− 1

x22

)= λ

∂g

∂x2

1

x1+

1

x2= m.

Dividing the first equation by the second givesv1

v2=x2

2

x21

.

As v1, v2, x1, x2,m are strictly positive we have

x2

x1=(v1

v2

) 12.

Substituting for x2 in the constraint occasion gives

1

x1+(v2

v1

) 12 · 1

x1= m

solving for x1 gives

1

x1

(1 + (

v1

v2)

12

)= m

x1 =(v1)

12 + (v2)

12

m · (v1)12

,

and similarly

x2 =(v1)

12 + (v2)

12

m · (v2)12

.

32. We want to optimizef(x1, x2) =

v1

v1 + v2· x1

2+

v2

v1 + v2· x2

2

subject to

g(x1, x2) =1

x1+

1

x2= m.


At the optimum point, x1, x2, and the Lagrange multiplier λ must satisfy the equations

v1

v1 + v2· 1

2= − λ

x21

v2

v1 + v2· 1

2= − λ

x22

1

x1+

1

x2= m.

Solving the first and second equations for 1/x1 and 1/x2, respectively, gives

1

x21

= − 1

2λ· v1

(v1 + v2)

1

x22

= − 1

2λ· v2

(v1 + v2)

substituting into the constraint gives (note that λ < 0):

(− 1

2λ· v1

(v1 + v2)

)1/2

+

(−1

2λ· v2

(v1 + v2)

)1/2

=(− 1

2λ

)1/2

· (v1)1/2 + (v2)1/2

(v1 + v2)1/2= m.

So

− 1

2λ· v1 + v2 + 2(v1v2)1/2

v1 + v2= m2.

and thus

λ = − 1

2m2

(1 +

2(v1v2)1/2

v1 + v2

).

The units of λ are weeks2 (since the units of m are 1/weeks). The Lagrange multiplier measures df/dm, whichrepresents the rate of change in the expected delay in tumor detection as the available number of visits per week increases.The negative sign represents the fact that as the number of visits per week increases, the delay decreases.

33. We want to minimize the total cost, C, of the cable. The distance AB is√

502 + x2 meters and the distance BC is√302 + y2, and the cost for each segment is cost/meter× length. Thus

C = 500√

502 + x2 + 2000√

302 + y2 + 300z,

where x, y, z must satisfy the constraint

g(x, y, z) = x+ y + z = 100.

At the optimum point,∇C = λ∇g, so

∂C

∂x=

500x√502 + x2

= λ = λ∂g

∂x

∂C

∂y=

2000y√302 + y2

= λ = λ∂g

∂y

∂C

∂z= 300 = λ = λ

∂g

∂z.

Thus, using λ = 300, we have

2000y√302 + y2

= 300

20y = 3√

302 + y2

400y2 = 9(900 + y2)

391y2 = 8100

y =90√391

= 4.6 meters.


Similarly

500x√502 + x2

= 300

5x = 3√

502 + x2

25x2 = 9(2500 + x2)

16x2 = 9 · 2500

x =3 · 50

4= 37.5 meters.

Thus z = 100− 37.5− 4.6 = 57.9 meters. The values x = 37.5 m, y = 4.6 m, and z = 57.9 m give the minimum cost.How do we know these values of x, y, z give a minimum? Since the variables are constrained to lie in the first octant

on the plane x + y + z = 100, we find the minimum cost on the boundary, which consists of three line segments on thecoordinate planes, and compare with the cost at the critical point.

If x = 0, then z = 100 − y and the minimum of C(x, y, z) = 500 · 50 + 2000√

302 + y2 + 300(100 − y) for0 ≤ y ≤ 100 is C(0, 0, 100) = $115,000.

If y = 0, then z = 100 − x and the minimum of C(x, y, z) = 500√

502 + x2 + 2000 · 30 + 300(100 − x) for0 ≤ x ≤ 100 is C(37.5, 0, 62.5) = $110,000.

If z = 0, then y = 100 − x and the minimum of C(x, y, z) = 500√

502 + x2 + 2000√

302 + (100− x)2 for0 ≤ x ≤ 100 is C(100, 0, 0) = $115,902.

The cost at the critical point,C(37.5, 4.6, 57.9) = $109,321, is lower than the minimum $110, 000 on the boundary.Thus, the minimum cost subject to the constraint is C(37.5, 4.6, 57.9) = $109,321.

34. The wetted perimeter of the trapezoid is given by the sum of the lengths of the three walls, so

p = w +2d

sin θ

We want to minimize p subject to the constraint that the area is fixed at 50 m2. A trapezoid of height h and with parallelsides of lengths b1 and b2 has

A = Area = h(b1 + b2)

2.

In this case, d corresponds to h and b1 corresponds to w. The b2 term corresponds to the width of the exposed surface ofthe canal. We find that b2 = w+ (2d)/(tan θ). Substituting into our original equation for the area along with the fact thatthe area is fixed at 50 m2, we arrive at the formula:

Area =d

2

(w + w +

2d

tan θ

)= d

(w +

d

tan θ

)= 50

We now solve the constraint equation for one of the variables; we will choose w to give

w =50

d− d

tan θ.

Substituting into the expression for p gives

p = w +2d

sin θ=

50

d− d

tan θ+

2d

sin θ.

We now take partial derivatives:

∂p

∂d= −50

d2− 1

tan θ+

2

sin θ∂p

∂θ=

d

tan2 θ· 1

cos2 θ− 2d

sin2 θ· cos θ

From ∂p/∂θ = 0, we getd · cos2 θ

sin2 θ· 1

cos2 θ=

2d

sin2 θ· cos θ.

Since sin θ 6= 0 and cos θ 6= 0, canceling gives1 = 2 cos θ

socos θ =

1

2.

Since 0 < θ <π

2, we get θ =

π

3.


Substituting into the equation ∂p/∂d = 0 and solving for d gives:

−50

d2− 1√

3+

2√3/2

= 0

which leads to

d =

√50√

3≈ 5.37m.

Thenw =

50

d− d

tan θ≈ 50

5.37− 5.37√

3≈ 6.21 m.

When θ = π/3, w ≈ 6.21 m and d ≈ 5.37 m, we have p ≈ 18.61 m.Since there is only one critical point, and since p increases without limit as d or θ shrink to zero, the critical point

must give the global minimum for p.

35. (a) A

B-� d

a

b

θ2

θ1

Medium 1

Medium 2

A′R

B′

Figure 15.39

See Figure 15.39. The time to travel from A to B is given by

T (θ1, θ2) =AR

v1+RB

v2=

a

v1 cos θ1+

b

v2 cos θ2.

(b) The distance d = A′B′ = A′R+RB′. Hence

d = a tan θ1 + b tan θ2.

(c) We imagine the following extreme case: the light ray first travels through medium 1 to a point R on the boundaryfar to the left of A′, then through medium 2 toward B. The distance traveled this way is very large, hence the traveltime is large as well. Similarly, if R is far to the right of B′, the travel time will be large. Therefore values of θ1 near−π/2 or π/2 increase the time, T .

(d) The constrained optimization problem is: minimize T (θ1, θ2) subject to g(θ1, θ2) = a tan θ1 + b tan θ2 = d.According to the method of Lagrange multipliers, the minimum point should be among those satisfying gradT =λ grad g as well as the constraint.We have

gradT (θ1, θ2) =a

v1

sin θ1

cos2 θ1

~i +b

v2

sin θ2

cos2 θ2

~j

andgrad g(θ1, θ2) = a

1

cos2 θ1

~i + b1

cos2 θ2

~j .

The condition gradT = λ grad g becomes

a

v1

sin θ1

cos2 θ1= λa

1

cos2 θ1and

b

v2

sin θ2

cos2 θ2= λb

1

cos2 θ2.

Eliminating λ we are left withsin θ1

v1=

sin θ2

v2

orsin θ1

sin θ2=v1

v2,

which is Snell’s law. The argument in part (c) shows that the critical point corresponding to θ1, θ2 satisfying Snell’slaw is indeed a minimum.


CAS Challenge Problems

36. (a) The partial derivatives of f are

∂f

∂x=

√a+ x+

(−1 +

√a+ x

)y

2√a+ x

√a+ x+ y

(1 +√a+ x+ y

)2

∂f

∂y=

1− 2 a− 2x+√a+ x− y

2√a+ x+ y

(1 +√a+ x+ y

)2

Solving ∂f/∂x = 0, ∂f/∂y = 0, we get x = 1/4 − a, y = 1. The discriminant at this point is D = −16/625.Thus, by the second derivative test, the point is a saddle point.

(b) The y coordinate of the critical points stays the same and the x coordinate is a units to the left of its position when

a = 0. The type is always a saddle point. This is because f is obtained from√x+ y

1 + y +√x

by substituting x + a for

x, so that the graph is shifted a units in the negative x-direction but its shape remains the same.

37. (a) We have grad f = 2x~i +~j and grad g = (2x+ 2y)~i + (2x+ 2y)~j . So the equations to be solved in the methodof Lagrange multipliers are

2x = λ(2x+ 2y)

1 = λ(2x+ 2y)

x2 + 2xy + y2 − 9 = 0

Solving these with a CAS, we get two solutions:

x = 1/2, y = −7/2, λ = −1/6, or x = 1/2, y = 5/2, λ = 1/6

Student A reasons that since f(1/2,−7/2) = −13/4 and f(1/2, 5/2) = 11/4 , the (global) maximum andminimum values are 11/4 and 13/4, respectively. Student B graphs the constraint curve g = 0 and a contour diagramof f . The constraint curve turns out to be two straight lines, since the constraint x2 + 2xy + y2 − 9 = 0, which canbe rewritten as (x+ y)2 = 9, or x+ y = ±3. The value of f goes to infinity on each of these straight lines. On theline y = −x+ 3, f(x, y) = x2 + y = x2 − x+ 3, and on the line y = −x− 3, f(x, y) = x2 + y = x2 − x− 3.Thus Student B is correct. The points Student A found are actually local maximum and local minimum values, notglobal. Since the constraint is not bounded, there is no guarantee that there is a local maximum or minimum. SeeFigure 15.40.

−4 −2 2 4

−4

−2

2

4

g

g

x

y

Figure 15.40: Contours of f and two straightlines giving constraint g = 0

38. (a) We have grad f = 3~i + 2~j and grad g = (4x− 4y)~i + (−4x+ 10y)~j , so the Lagrange multiplier equations are

3 = λ(4x− 4y)

2 = λ(−4x+ 10y)

2x2 − 4xy + 5y2 = 20


Solving these with a CAS we get λ = −0.4005, x = −3.9532, y = −2.0806 and λ = 0.4005, x = 3.9532, y =2.0806. We have f(−3.9532,−2.0806) = −11.0208, and f(3, 9532, 2.0806) = 21.0208. The constraint equationis 2x2 − 4xy+ 5y2 = 20, or, completing the square, 2(x− y)2 + 3y2 = 20. This has the shape of a skewed ellipse,so the constraint curve is bounded, and therefore the local maximum is a global maximum. Thus the maximum valueis 21.0208.

(b) The maximum value on g = 20.5 is ≈ 21.0208 + 0.5(0.4005) = 21.2211. The maximum value on g = 20.2 is≈ 21.0208 + 0.2(0.4005) = 21.1008.

(c) We use the same commands in the CAS from part (a), with 20 replaced by 20.5 and 20.2, and get the maximumvalues 21.2198 for g = 20.5 and 21.1007 for g = 20.2. These agree with the approximations we found in part (b) to2 decimal places.

CHECK YOUR UNDERSTANDING

1. True. By definition, a critical point is either where the gradient of f is zero or does not exist.

2. False. The point P0 could be a saddle point of f .

3. False. The point P0 could be a saddle point of f .

4. True. If P0 were not a critical point of f , then grad f(P0) would point in the direction of maximum increase of f , whichcontradicts the fact that P0 is a local maximum or minimum.

5. True. The graph of this function is a cone that opens upward with its vertex at the origin.

6. False. The graph of this function is a saddle shape, with a saddle point at the origin. The function increases in the ~idirection and decreases in the ~j direction.

7. True. Adding 5 to the function shifts the graph 5 units vertically, which leaves the (x, y) coordinates of the local extremaintact.

8. True. Multiplying by −1 turns the graph of f upside down, so local maxima become local minima and vice-versa.

9. False. For example, the linear function f(x, y) = x+ y has no local extrema at all.

10. False. The statement is only true for points sufficiently close to P0.

11. False. Local maxima are only high points for f when compared to nearby values; the global maximum is the largest ofany values of f over its entire domain.

12. True. For unconstrained optimization, global extrema occur at one (or more) of the local extrema.

13. False. For example, the linear function f(x, y) = x + y has neither a global minimum or global maximum on all of2-space.

14. True. The region is the unit disk without its boundary (the unit circle), and the distance between any two points in thisregion is less than 2—it does not stretch off to infinity in any direction.

15. False. The region is the unit disk without its boundary (the unit circle), so it is not closed (in fact, it is open).

16. True. The global minimum is 0, which occurs at the origin. This is clear since the function f(x, y) = x2 + y2 is greaterthan or equal to zero everywhere, and is only zero at the origin.

17. False. On the given region the function f is always less than one. By picking points closer and closer to the circlex2 + y2 = 1 we can make f larger and larger (although never larger than one). There is no point in the open disk thatgives f its largest value.

18. False. While f can only have (at most) one largest value, it may attain this value at more than one point. For example, thefunction f(x, y) = sin(x+ y) has a global maximum of 1 at both (π/2, 0) and (0, π/2).

19. True. The region is both closed and bounded, guaranteeing both a global maximum and minimum.

20. True. The global minimum could occur on the boundary of the region.

21. True. The point (a, b) must lie on the constraint g(x, y) = c, so g(a, b) = c.

22. False. The point (a, b) is not necessarily a critical point of f , since it is a constrained extremum.

23. True. The constraint is the same as x = y, so along the constraint f = 2x, which grows without bound as x→∞.

24. False. The condition grad f = λgrad g yields the two equations 1 = λ2x and 2 = λ2y. Substituting x = 2 in the firstequation gives λ = 1/4, while setting y = −1 in the second gives λ = −1, so the point (2,−1) is not a local extremumof f constrained to x+ 2y = 0.

25. False. Since grad f and grad g point in opposite directions, they are parallel. Therefore (a, b) could be a local maximumor local minimum of f constrained to g = c. However the information given is not enough to determine that it is a

PROJECTS FOR CHAPTER FIFTEEN 1101

minimum. If the contours of g near (a, b) increase in the opposite direction as the contours of f , then at a point withgrad f(a, b) = λgrad g(a, b) we have λ ≤ 0, but this can be a local maximum or minimum.

For example, f(x, y) = 4 − x2 − y2 has a local maximum at (1, 1) on the constraint g(x, y) = x + y = 2. Yet atthis point, grad f = −2~i − 2~j and grad g =~i +~j , so grad f and grad g point in opposite directions.

26. False. A maximum for f subject to a constraint need not be a critical point of f

27. False. The condition for the Lagrange multiplier λ is grad f(a, b) = λ grad g(a, b).

28. False. Just as a critical point need not be a maximum or minimum for unconstrained optimization, a point satisfying theLagrange condition need not be a maximum or minimum for a constrained optimization.

29. True. Since f(a, b) = M , we must satisfy the Lagrange conditions that fx(a, b) = λg(a, b) and fy(a, b) = λgy(a, b),for some λ. Thus fx(a, b)/fy(a, b) = gx(a, b)/gy(a, b).

30. True. Since f(a, b) = m, the point (a, b) must satisfy the Lagrange condition that fx(a, b) = λg(a, b), for some λ. Inparticular, if gx(a, b) = 0, then fx(a, b) = 0.

31. False. Whether increasing c will increase M depends on the sign of λ at a point (a, b) where f(a, b) = M

32. True. The value of λ at a maximum point gives the proportional change in M for a change in c.

33. False. The value of λ at a minimum point gives the proportional change in m for a change in c. If λ > 0 and the changein c is positive, the change in m will also be positive.

PROJECTS FOR CHAPTER FIFTEEN

1. (a) If p = e−x where x→∞ then p→ 0 with p > 0 and

limp→0p>0

(p ln p) = limx→∞

(−xe−x) = 0,

since the exponential decreases faster than any power of x. Alternatively, use l’Hopital’s rule:

limp→0p>0

(p ln p) = limp→0p>0

ln p

1/p= lim

p→0p>0

1/p

−1/p2= 0

(b) We apply the method of Lagrange multipliers to find the critical points of S(p1, · · · , p30). The constraintfunction is g(p1, · · · , p30) = p1 + · · ·+ p30. We have

∂S

∂pj=

∂

∂pj

(−

30∑

i=1

piln piln 2

)= − 1

ln 2(ln pj + 1),

therefore

gradS = − 1

ln 2

30∑

j=1

(ln pj + 1)~kj

where ~k1 , · · · , ~k30 are the unit vectors corresponding 30 independent directions of the pj-axes. Also,

grad g =30∑

j=1

~kj

so the condition gradS = λ grad g becomes

− 1

ln 2(ln pj + 1) = λ, for i = 1, · · · , 30.

Thus,ln pj = −λ ln 2− 1


and, in particular, all the pjsmust be equal. Since the pjs have to satisfy the constraint g(p1, · · · , p30) = 1,we see that pj = 1

30 and that the point ( 130 ,

130 , · · · , 1

30 ) is the only critical point of S. We have

S

(1

30,

1

30, · · · , 1

30

)= −30 · 1

30

(− ln 30)

ln 2=

ln 30

ln 2.

We will not prove that this is indeed the maximum value of S (this requires a higher-dimensional analogueof the second derivative test). Since in part (c) we show that the minimum value of S is 0, the critical pointwe have found here is not a global minimum; the maximum of S has to be attained somewhere and it isreasonable to believe that it is attained at the unique critical point. The maximum entropy corresponds tomaximum uncertainty in the outcome of the competition.

(c) We already know that S ≥ 0. However, S can be zero: For example, if p1 = 1 and p2 = · · · = p30 = 0,we have S(1, 0, · · · , 0) = 0. Therefore the minimum value of S is 0. Now we determine all the values ofpis for which S(p1, · · · , p30) = 0. The condition

S = −30∑

i=1

pi ln pi2

= 0

together with the restrictions − ln pi ≥ 0 shows that, for S to vanish, each individual term in the abovesum has to vanish. This means pi ln pi = 0 for all i = 1, · · · , 30, that is, pi = 0 or pi = 1 for i = 1, · · · , 30.Since

∑30i=1 pi = 1, only one of the pis is 1 whereas the other 29 are 0. This corresponds to the case where

one of the teams is certain to win, that is, there is no uncertainty. The result can be interpreted by sayingthat zero entropy implies zero uncertainty.

2.

(xi, b+mxi)

(xi, yi)(x1, b+mx1)

(x1, y1)

y = b+mx

Figure 15.41

(a) Points which are directly above or below each other share the same x coordinate, therefore, the point onthe least squares line which is directly above or below the point in question will have x coordinate xi andfrom the formula for the least squares line, it will have y coordinate b+mxi. (See Figure 14.1.)

(b) The general distance formula in two dimensions is d =√

(x2 − x1)2 + (y2 − y1)2, so d2 = (x2−x1)2 +(y2 − y1)2. Since the x coordinates are identical for the two points in question, the first term in the squareroot is zero. This yields d2 = (yi − (b+mxi))

2.(c) In both cases we use the chain rule and our knowledge of summations to show the relationship.

∂f

∂b=

∂

∂b

(n∑

i=1

(yi − (b+mxi))2

)=

n∑

i=1

∂

∂b(yi − (b+mxi))

2

=

n∑

i=1

2(yi − (b+mxi)) ·∂

∂b(yi − (b+mxi))

=

n∑

i=1

2(yi − (b+mxi)) · (−1)

= −2

n∑

i=1

(yi − (b+mxi))

PROJECTS FOR CHAPTER FIFTEEN 1103

∂f

∂m=

∂

∂m

(n∑

i=1

(yi − (b+mxi))2

)=

n∑

i=1

∂

∂m(yi − (b+mxi))

2

=n∑

i=1

2(yi − (b+mxi)) ·∂

∂m(yi − (b+mxi))

=

n∑

i=1

2(yi − (b+mxi)) · (−xi)

= −2n∑

i=1

(yi − (b+mxi)) · xi

(d) We can separate∂f

∂binto three sums as shown:

∂f

∂b= −2

(n∑

i=1

yi − bn∑

i=1

1−mn∑

i=1

xi

)

Similarly we can separate∂f

∂mafter multiplying through by xi:

∂f

∂m= −2

(n∑

i=1

yixi − bn∑

i=1

xi −mn∑

i=1

xi2

)

Setting∂f

∂band

∂f

∂mequal to zero we have:

bn+m

n∑

i=1

xi =

n∑

i=1

yi

b

n∑

i=1

xi +m

n∑

i=1

x2i =

n∑

i=1

xiyi

(e) To solve this pair of linear equations, we multiply the first equation by∑ni=1 x

2i , multiply the second one

by∑ni=1 xi, and subtract; we get

bn

n∑

i=1

x2i − b(

n∑

i=1

xi)2 =

n∑

i=1

yi

n∑

i=1

x2i −

n∑

i=1

xiyi

n∑

i=1

xi,

So,

b =

(n∑

i=1

x2i

n∑

i=1

yi −n∑

i=1

xi

n∑

i=1

xiyi

)/

n

n∑

i=1

x2i −

(n∑

i=1

xi

)2

Similarly,

m =

(n

n∑

i=1

xiyi −n∑

i=1

xi

n∑

i=1

yi

)/

n

n∑

i=1

x2i −

(n∑

i=1

xi

)2

(f) Applying the formulas to the given data, we have b = − 13 , m = 1 which gives y = −(1/3) + x, in

agreement with the example.

ch15 (1)

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f1 f2 mg

2h 4r

bx 2cy

lagrangian

low daily

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partial derivatives

maximumand