2. economic optimization

7/16/2019 2. Economic Optimization

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LESSON TWO : ECONOMIC OPTIMIZATION/BASIC TRAINING

A. Methods of Differentiation

The slope of a function y = f(x) is the change in y divided by the correspondingchange in x. In order to find the value of slope, there is a need to use derivatives.

For a function y=f(x), the derivative, written as dy/dx is the slope of the function ata particular point on the function.

Rules of differentiation:

1. The Derivative of a constant.

The derivative of any constant is always zero.

Example:y = 10dy/dx = 0

2. The derivative of a power function.

If y = axn then, dy/dx = n*ax (n – 1)

Example:y = 10x2

dy/dx = (2)10x(2-1) = 20x

3. The derivative of a constant times a function.

If y = a*f(x), where x is constant, dy/dx is a.

Example:

y = 10 x(1-1)

dy/dx = (1)10X(0) = 10

4. The derivative of a sum or difference.

If y = f(x) + g(x), then dy/dx = f’(x) + g’(x).

Example:y = 20x + 10dy/dx = (1)20x(1-1) + 0 = 20

5. The derivative of a product function.

If y = f(x)g(x), then dy/dx = f’(x)g(x) + f(x)g’(x)

Where f’(x) = df/dx and g’(x) =dg/dx

Example:y = (x2 – 3)(x3 + 4x + 2)

Assume that f(x) = (x2 – 3) and g(x) = (x3 + 4x + 2)

f’(x) = df/dx = 2x and g’(x) =dg/dx =(3x2 + 4)

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so to find dy/dx , we use the following formula:

dy/dx = f’(x)g(x) + f(x)g’(x)

Thus, dy/dx = (2x)(x3 + 4x + 2) + (x2 – 3)(3x2 + 4)

= [2x4 + 8x2 + 4x] + [3x4 + 4x2 – 9x2 + 12]

dy/dx = 5x4 + 3x2 + 4x + 12

If the value of x is given (x=1) thendy/dx = 5(1)4 + 3(1)2 + 4(1) + 12

= 24

6. The derivative of a quotient function.

If y = f(x)/g(x), then

dy/dx = g(x)f’(x) – f(x)g’(x)[g(x)]2

Example:y = x2 -4x

x2

dy/dx = x2(2x-4) – (x2 – 4x)(2x)

(x2)2

= (2x3 – 4x2) – (2x3 – 8x2)

(x4)

= 2x3 – 4x2 – 2x3 + 8x2

(x4)

= 4x2

x4

dy/dx = 4/x2

Exercise 1A:

Determine the derivatives of each of the following functions.

a) Y = 20

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b) Y = 6X2 + 8X + 50

c) Y = (X2 - 2)(X2 + 4X +10)

d) Y = X2 + 3X + 4

X

2

– 4

e) Q = 120 – 0.2P2

f) C = 2,500 – 100X2 + 4X3

B. First and Second Derivatives

i) The maximum or minimum point of a function, y = f(x) can be found by settingthe first derivative of the function to zero and solving the equation for thevalue(s) of x.

Example 1B:

If y = 200x – 0.4x2, find the optimum value of x.Find dy/dx = 0 allows you to find the optimum value of x.

dy/dx = 200 – 0.8x = 00.8x = 200X = 200/0.8x = 25

ii) When the first derivative of a function is zero, the function is at a maximum if the second derivative is negative, or at a minimum if the second derivativeis positive.

From example 1B, the first derivative of y = 200x – 0.4x2

is dy/dx = 200 – 0.8x.Setting it to zero, the optimum value of x is 25. To determine whether thisvalue of x is minimum or maximum, you have to find the second derivative of the function.

Since, dy/dx = 200 – 0.8xd2y/dx2 = -0.8

since d2y/dx2 is less than zero, then x = 25 is the maximumvalue.

Example 2B:

If total revenue, TR = 200Q – 2Q2 , what is the output level that can maximizeTR?

Maximize TR = dTR = MR = 0

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dQ= 200 – 4Q = 0

4Q = 200Q = 50

Thus, in order to maximize TR, output should be equal to 50 units.

To determine whether Q = 50 is indeed the maximum value, you haveto find the second derivative of the TR function.

dTR/dQ = 200 – 4Qd2 TR/dQ2 = -4 (less than zero)

This shows that Q = 50 is indeed a maximum value.

iii. Sometimes, setting the first derivatives to zero will give two extremum values.In order to identify which one is the optimum value, one has to substitute thetwo values into the second derivative.

Example 3B:

TR = 7Q – 0.1Q2

TC = 10 + 8Q – 0.3Q2 + 0.01Q3

Find the value of Q that will maximize profit.

Profit = TR – TC= -10 – Q + 0.2Q2 – 0.01Q3

d(Profit) = -1 + 0.4Q – 0.03Q2 = 0dQ

= (3Q – 10) (Q – 10) = 0

Thus, 3Q – 10 = 0 OR Q – 10 = 0Q = 10/3 Q = 10

There should be only one optimum value. Therefore, substitute thesevalues into the second derivative.

d2 (Profit) = 0.4 – 0.06QdQ2

substitute Q = 10/3 into the second derivative

d2 (Profit) = 0.4 – 0.06(10/3)dQ2

= 0.2 (positive value = minimum)

d2 (Profit) = 0.4 – 0.06(10)dQ2

= -0.2 (negative value = maximum)

The value of Q that will maximize profit is Q = 10.

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iv. To be an optimal value, the value must fulfill both first order condition(setting first derivative equals zero) and second order condition(finding the second derivative.)

Exercise1B:

1. Optimize the following functions and find the value of x at the optimumpoint. Then determine whether x is a minimum or maximum point.a) y = 3x2 – 4x + 25

b) y = (3x – 2 )2

2. Given the following total revenue and total cost functions of a firm:

TR = 22Q – 0.5Q2

TC = 1/3Q3 – 8.5Q2 + 50Q + 90

determine,

a) the level of output at which the firm maximizes its total profitb) the maximum profit that the firm could earn.

C. Partial Derivative

In the previous section, the exercises given involve differentiation of simple functioni.e. one dependent and one independent variables. However, functions are normallyin the form of multivariate functions i.e. one dependent but many independentvariables. In order to find the derivative of the multivariate function, we have to findthe partial derivative of the function.

Example:

y = 10 - 8x

2

- 130z + xz + 2z

2

i) Find the partial derivatives of the function with respect to eachindependent variable x and z and set these derivatives

equal to zero.

dy/dx = - 16x + z = 0 ---------- (1)

dy/dz = -130 + x + 4z = 0 --------- (2)

ii) Solve the two equations simultaneously for X and Z.

From (1), Substitute z = 16x in (2) gives

-130 + x + 4(16x) = 0-130 + x + 64x = 0-130 + 65x = 0

65x = 130x = 130/65x = 2

Substitute x = 2 in z = 16x

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z = 16x= 16 (2)

z = 32

Exercise 1C:

1. Find the partial derivative for the following functions.

a) y = 4 – x2 – 2z + xz + 2z2

b) AC = x2 + 2y2 – 2xy – 2x -6y + 20

2. For the following total-profit function of a firm:

= 144X – 3X2 – XY – 2Y2 + 120Y - 35

determine;

i) the level of output of each commodity at which the firmmaximizes its total profit,

ii) the value of the maximum amount of the total profit of the firm

3. For the following function,

Y = 50 + 18X + 10Z – 5XZ – 2X2

find the value of X and Z that will maximize the function. What is themaximum value of Y?

D. Lagrangian : Solving a constrained optimization problem

In many decision problems, there are constraints imposed that limit the options of choicesavailable to the decision makers.

Example:

TC = 3x2 + 6y2 – xy

x = output produced by line 1y = output produced by line 2

Objective function : TC = 3x2 + 6y2 – xyConstraint function : x + y = 20

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Methods:

1. normal/substitution method2. lagrangian method

Method 1: Normal/substitution method

TC = 3x2 + 6y2 – xy ------ (1)x + y = 20 ------ (2)X = 20 – y

Substitute (2) into (1): TC = 3(20 – y)2 + 6y2 – (20 – y)y TC = 1200 – 140y + 10y2 ---------- (3)

d(TC)/dy = -140 + 20y = 0y = 7 -------- (4)

Substitute (4) into (2) will give x = 13 -------- (5)

Substitute (4) and (5) into (1) will give TC = 710

Thus, at x = 13 and y = 7, total cost is minimized at 710.

Method 2 : lagrangian

Lagrangian is a process to combine the objective function and the constraint functionwith a multiplier (lambda).

Steps:(i) Objective function : TC = 3x2 + 6y2 – xy(ii) Constraint function : x + y = 20(iii) Rewrite the constraint function as follows:

x + y – 20 = 0OR

20 – x – y = 0

(iv) Lagrangian function :

L TC = 3x2 + 6y2 – xy - λ(x + y – 20)OR

L TC = 3x2 + 6y2 – xy + λ(20 – x - y)

(v) Take all partial derivatives and set equal to zero

(vi) Solve simultaneous equation(v) Interpret lambda

Solution:

L TC = 3x2 + 6y2 – xy – λ(x + y – 20)L TC = 3x2 + 6y2 – xy – xλ – yλ + 20λ

dL TC /dx = 6x – y – λ = 0 ------ (1)dL TC /dy = 12y – x – λ = 0 ------ (2)

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dL TC /dλ = – x – y + 20 = 0 ------ (3)

(1) – (2) : 7x – 13y = 0 ------ (4)(4) : x = 13y/7 ------ (5)Sub.(5) into (3) : –(13y/7) – y + 20 = 0

[–13y – 7y + 140] = 0

-------------------------7

[–13y – 7y + 140] = 0 ×7

– 20y – 140 = 0

Y = 7 ------ (6)

Substitute (6) into (4):7x – 13y = 07x – 13(7) = 07x – 91 = 0X = 91/7

x = 13 ------ (7)

Substitute (6) and (7) into (1):6(13) – 7 – λ = 078 – 7 – λ = 071 – λ = 0

λ = 71

Substitute (6) and (7) into TC function; TC = RM710

λ = 71 means if output (constraint) were to increase by one unit from 20 to 21, thenthe TC (obj. function) will increase by 71.

Exercise 1D:

1. Firm AMC total profit function is given as

= 80x – 2x2 – xy – 3y2 + 100y

It faces the constraint that the output of commodity x plus output of commodity y must be 12.

a. Use the Lagrangian Multiplier method to find profit maximization.b. Interpret the value of lambda.

2. Oct 2003 Part A Question 2

3. April 2009 Part A Question 2

4. Oct 2009 Part A Question 6

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2. economic optimization

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