Optimization and Derivatives

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


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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


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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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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



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




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



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.



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



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.



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.



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



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


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

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Optimization and Derivatives