LESSON TWO : ECONOMIC OPTIMIZATION/BASIC TRAINING


 PBUY


 


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 corresponding change 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 at a particular point on the function.

Rules of differentiation:

1.         The Derivative of a constant.

            The derivative of any constant is always zero.

            Example:
                        y          =          10
                        dy/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 + 10
                        dy/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)

           
            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) then
                        dy/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

b)                 Y = 6X2 + 8X + 50

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

d)                 Y = X2 + 3X + 4
           X2 – 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 setting the first derivative of the function to zero and solving the equation for the value(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        = 0
                                                             0.8x                 = 200
                                                             X                     = 200/0.8
                                                             x                      = 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 derivative is 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 this value of x  is minimum or maximum, you have to find the second derivative of the function.

                        Since,   dy/dx               = 200  – 0.8x
                                    d2y/dx2             = -0.8

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

                       


            Example 2B:

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

                        Maximize TR = dTR        = MR = 0
                                                dQ
                                                            = 200 – 4Q        = 0
                                                                        4Q        = 200
                                                                        Q         = 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 have to find the second derivative of the TR function.
            dTR/dQ            = 200 – 4Q
            d2TR/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 the two 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           = 0
                 dQ
                                    = (3Q – 10) (Q – 10)     = 0
           
            Thus,     3Q – 10 = 0                   OR                   Q – 10  = 0
                        Q         = 10/3                                      Q         = 10

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

            d2 (Profit)         =          0.4 – 0.06Q
                  dQ2

            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.

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 optimum point. 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 profit
                        b)         the maximum profit that the firm could earn.

C.         Partial Derivative

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

Example:

                                    y          =          10 - 8x2 - 130z + xz + 2z2
           
            i)          Find the partial derivatives of the function with respect to each                                independent 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           = 130
                                                  x           = 130/65
                                                  x           =  2

                        Substitute x = 2  in  z = 16x
                                   
                                    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 firm       maximizes 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 the                                     maximum value of Y?








D.         Lagrangian : Solving a constrained optimization problem

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

Example:

            TC        =          3x2 + 6y2 – xy

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

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

Methods:

            1.         normal/substitution method
            2.         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         =          0
                                                            y          =          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 function with 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 = 0
            OR
                        20 – x – y = 0   
           
(iv)       Lagrangian function    :          
                                               
                                                LTC = 3x2 + 6y2 – xy -  λ(x + y – 20)
                                    OR
                                                LTC = 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:

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

dLTC /dx            =            6x – y – λ        = 0       ------      (1)
dLTC /dy           =          12y – x –  λ       = 0       ------      (2)
dLTC /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 = 0
            7x – 13(7) = 0
            7x – 91     = 0
            X = 91/7

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

Substitute (6) and (7) into (1):
            6(13) – 7 – λ     = 0
            78 – 7 – λ          = 0
            71 – 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, then the 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





life priced,do not fight,peace no war

Tiada ulasan:

Catat Ulasan