Oct 19, 2015
5/28/2018 Optimization and Derivatives
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Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Managerial Economics
Optimization Techniques
and New Management Tools
5/28/2018 Optimization and Derivatives
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Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
OPTIMIZATION
Managerial economics is concerned with the ways in which
managers should make decisionsin order to maximize the
effectiveness or performance of the organizations theymanage. To understand how this can be done we must
understand the basic optimization techniques.
Functional relationships:relationships canbe expressed by graphs:
P
Q
5/28/2018 Optimization and Derivatives
3/13
Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Quick Differentiation Review
Constant Y = c dY/dX = 0 Y = 5
Functions dY/dX = 0
A Line Y = c X dY/dX = c Y = 5X
dY/dX = 5
Power Y = cXb dY/dX = bcX b-1 Y = 5X2
Functions dY/dX = 10X
Name Function Derivative Example
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Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Sumof Y = G(X) + H(X) dY/dX = dG/dX + dH/dXFunctions
example Y = 5X + 5X2 dY/dX = 5 + 10X
Product of Y= G(X) H(X)
Two Functions dY/dX = (dG/dX)H + (dH/dX)G
example Y = (5X)(5X2)
dY/dX = 5(5X2) + (10X)(5X) = 75X2
Quick Differentiation Review
5/28/2018 Optimization and Derivatives
5/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Quotient of Two Y = G(X) / H(X)Functions
dY/dX = (dG/dX)H - (dH/dX)GH2
Y = (5X) / (5X2) dY/dX = 5(5X2) -(10X)(5X)(5X2)2
= -25X2/ 25X4= - X-2
Chain Rule Y = G [ H(X) ]
dY/dX = (dG/dH)(dH/dX) Y = (5 + 5X)2
dY/dX = 2(5 + 5X)1(5) = 50 + 50X
Quick Differentiation Review
5/28/2018 Optimization and Derivatives
6/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
USING DERIVATIVES TO SOLVE MAXIMIZATION AND
MINIMIZATION PROBLEMS
Maximum or minimum points occur only if the slope of the
curve equals zero.
Look at the following graph
5/28/2018 Optimization and Derivatives
7/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
y
dy/dx 10
10 20
20
0
Max of x
Slope = 0
value of x
Value of dy/dx whichIs the slope of y curve
Value of Dy/dx when y is max
x
x
the function
Y = -50 + 100X - 5X2
i.e.,
dY
dX= 100 - 10X
ifdY
dX= 0
X = 10
i.e., Y is maximized when
the slope equals zero.
Note that this is not sufficient for maximization or minimization problems.
5/28/2018 Optimization and Derivatives
8/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Optimization Rules
Maximization conditions:
1 -dY
dX= 0
2 -d Y
dX
2
2
= < 0
Minimization conditions:
1 - dYdX
= 0
2 -d Y
dX
2
2= > 0
5/28/2018 Optimization and Derivatives
9/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Applications of Calculus in Managerial Economics
maximization problem:
A profit function might look like an arch, rising to a peak and thendeclining at even larger outputs. A firm might sell huge amountsat very low prices, but discover that profits are low or negative.
At the maximum, the slope of the profit function is zero. Thefirstorder condition for a maximum is that the derivative at that pointis zero.
If = 50Q - Q2,
then d/dQ = 50 - 2Q, using the rules of differentiation.Hence, Q = 25 will maximize profits
where
50 - 2Q = 0.
5/28/2018 Optimization and Derivatives
10/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
More Applications of Calculus
minimization problem: Cost minimization
supposes that there is a least cost point to produce. Anaverage cost curve might have a U-shape. At the leastcost point, the slope of the cost function is zero. The
first order conditionfor a minimum is that the
derivative at that point is zero.
If TC = 5Q260Q,
then dC/dQ = 10Q - 60.
Hence, Q = 6will minimize cost
Where:
10Q - 60 = 0.
5/28/2018 Optimization and Derivatives
11/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
More Examples
Competitive Firm: Maximize Profits
where = TR - TC = P Q - TC(Q)
Use our first order condition:
d /dQ = P - dTC/dQ = 0. Decision Rule: P = MC.
a function of Q
Max = 100Q - Q2
First order = 100 -2Q = 0 implies
Q = 50 and;
= 2,500
5/28/2018 Optimization and Derivatives
12/13Managerial Economics Prof. M. El-Sakka CBA. Kuwait University
Second Order Condition: one variable
If the second derivative is negative, then its a maximum
If the second derivative is positive, then its a minimum
Max = 100Q - Q2
First derivative
100 -2Q = 0
second derivative is: -2implies
Q =50 is a MAX
Max = 50 + 5X2
First derivative
10X = 0
second derivative is: 10implies
Q = 10 is a MIN
Problem 1 Problem 2
5/28/2018 Optimization and Derivatives
13/13Managerial Economics Prof M El-Sakka CBA Kuwait University
e.g.;
Y = -1 + 9X - 6X
2
+ X
3
first condition
dY
dX= 9 - 12X + 3X
2= 0
Quadratic Function
Y = aX2+ bX + c
X = b b ac
a
2 4
2
a = 3
b = -12
c = 9
Extra examples
X = ( ) ( )12 122 4 9 3
6
= 2 1
therefore
Y = 0 at
X = 3 or X = 1
the second condition
d Y
dX
22
= -12 + 6X
at X = 3
d Y
dX
2
2
= -12 + 6(3) = 6 >0 ( minimum point)
at X = 1
d Y
dX
2
2= -12 + 6(1) = - 6