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
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
3x^2+6x^2-xy
BalasPadam