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Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17
14

Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Dec 20, 2015

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Page 1: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Chapter One Homeworkdue tomorrow in lab

Numbers 10, 12 and 17

Page 2: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Managerial Economics & Business Strategy

Chapter 1The Fundamentals of Managerial

Economics

Page 3: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Net Benefits

• Net Benefits = Total Benefits - Total Costs

• Profits = Revenue - Costs

Page 4: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Marginal Benefit (MB)

• Change in total benefits arising from a change in the control variable, Q:

• Slope (first derivative) of the total benefit curve.

Q

BMB

Page 5: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Marginal Cost (MC)

• Change in total costs arising from a change in the control variable, Q:

• Slope (first derivative) of the total cost curve

Q

CMC

Page 6: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Marginal Principle

• To maximize net benefits MB = MC.• MB > MC means the last unit of the control

variable increased benefits more than it increased costs.

• MB < MC means the last unit of the control variable increased costs more than it increased benefits.

• Can we do it??? Start with a control variable that can only be used in WHOLE units (discrete)

Page 7: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Control Variable

Total Benefits

Total Costs

Net Benefits

Marginal Benefits

Marginal Costs

Marginal Net

Benefits

Q B(Q) C(Q) N(Q) MB(Q) MC(Q) MNB(Q)

0 0 0

1 90 10

2 170 30

3 240 60

4 300 100

5 350 150

6 390 210

7 420 280

8 440 360

9 450 450

10 450 550

Page 8: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Control Variable

Total Benefits

Total Costs

Net Benefits

Marginal Benefits

Marginal Costs

Marginal Net

Benefits

Q B(Q) C(Q) N(Q) MB(Q) MC(Q) MNB(Q)

0 0 0 0

1 90 10 80

2 170 30 140

3 240 60 180

4 300 100 200

5 350 150 200

6 390 210 180

7 420 280 140

8 440 360 80

9 450 450 0

10 450 550 -100

Page 9: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Control Variable

Total Benefits

Total Costs

Net Benefits

Marginal Benefits

Marginal Costs

Marginal Net

Benefits

Q B(Q) C(Q) N(Q) MB(Q) MC(Q) MNB(Q)

0 0 0 0 ---

1 90 10 80 90

2 170 30 140 80

3 240 60 180 70

4 300 100 200 60

5 350 150 200 50

6 390 210 180 40

7 420 280 140 30

8 440 360 80 20

9 450 450 0 10

10 450 550 -100 0

Page 10: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Control Variable

Total Benefits

Total Costs

Net Benefits

Marginal Benefits

Marginal Costs

Marginal Net

Benefits

Q B(Q) C(Q) N(Q) MB(Q) MC(Q) MNB(Q)

0 0 0 0 --- ---

1 90 10 80 90 10

2 170 30 140 80 20

3 240 60 180 70 30

4 300 100 200 60 40

5 350 150 200 50 50

6 390 210 180 40 60

7 420 280 140 30 70

8 440 360 80 20 80

9 450 450 0 10 90

10 450 550 -100 0 100

Page 11: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Control Variable

Total Benefits

Total Costs

Net Benefits

Marginal Benefits

Marginal Costs

Marginal Net

Benefits

Q B(Q) C(Q) N(Q) MB(Q) MC(Q) MNB(Q)

0 0 0 0 --- --- ---

1 90 10 80 90 10 80

2 170 30 140 80 20 60

3 240 60 180 70 30 40

4 300 100 200 60 40 20

5 350 150 200 50 50 0

6 390 210 180 40 60 -20

7 420 280 140 30 70 -40

8 440 360 80 20 80 -60

9 450 450 0 10 90 -80

10 450 550 -100 0 100 -100

Page 12: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

What if we can use fractional parts of the control variable?

Q

Total Benefits & Total Costs

Benefits

Costs

Q*

B

CSlope = MC

Slope =MB

Page 13: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

Let’s talk calculus…• What is the MB?

300-12Y

• What is the MC? 8Y

• What is the profit max level of Y? 300-12Y = 8Y 300=20Y Y=15

• What are the Net Benefits at this level of Y?

NB = ((300*15)-(6*15^2))-(4*15^2) NB = 2250

2

2

4)(

6300)(

YYC

YYYB

Page 14: Chapter One Homework due tomorrow in lab Numbers 10, 12 and 17.

One more check….Is it a maximum point??

• Second derivative of the Net Benefit equation must be negative

• N(Y) = B(Y)-C(Y)• First derivative

(300-12Y)-8Y

• Second derivative -12-8= -20 Negative Maximum

• Shows N(Y) is concave Slope of MB curve < slope of MC curve

2

2

4)(

6300)(

YYC

YYYB