According to chain rule if Y = F(U) and U = g(X), the derivative of Y with respect to X can be procured by multiplying mutually the derivative of Y with respect to U and the derivative of U with respect to X.
Let us see an illustration of this type.
Illustration 5
Presume Y = U^3 + 30 and U = 6X^2. Compute using the Chain rule.
Solution:
dY = dY * dU
dX dU dX
= d(U^3
+ 30) * d(6X^2)
dU dX
Derivative of U^3 + 30 = 3*U^3-2 = 3U^2
Derivative of 6X^2 = 12X^2-1 = 12X^1 = 12X
Derivative of dU and dX = 1
= 3U^2 * 12X
Substituting U = 6X^2 in the above, we get the following.
dY = 3(6X^2)^2
* 12X
dX
= 18X^4 * 12X
= 216X^5
Application of Derivatives to Optimisation Problems
Use in Profit Maximisation
Illustration 6
The profit function is as follows: π = -50 + 80Q – 20Q^2. π is the profit and Q is the units of productivity. Compute the derivative function setting it to zero.
Solution
dπ = 80 – 40Q
dQ
80 – 40Q = 0
40Q = 80
Q = 80 / 40
Q = 2.
Hence, at 8 units of productivity profits will be optimum.
Substituting the value of Q = 2 in the primary function gives the following.
π = -50 + 80(2) – 20(2)^2
= -50
+ 160 – 20*4
= 110 – 80
= 30
Therefore, the productivity level of 2 units, profits are equal to 30.
Minimisation Problem
Illustration 7
Let us assume the average function of a firm is given as below:
AC = 10,000 – 90Q + 0.5Q^2
Ascertain the level of productivity to minimise the average cost.
Solution
To derivate the Average Cost Function with respect to productivity Q is attained by parity to zero. Therefore,
d(AC) = -90
+ 1Q
d(Q)
Defining the equation by setting to zero procures,
-90 + Q = 0
Hence, Q = 90
By applying the second condition to make certain whether it is really minimum, we consider the second derivation of AC function
d^2AC = 1
dQ^2
As the second condition derivative AC function is positive and d^2AC > 0,
dQ^2
productivity of 90 units of output is the one that minimises the average cost
of manufacturing.
Multivariate Maximisation
Illustration 8
Let us assume the following two product function of a firm. π = 100X – 4X^2 – XY – 8Y^2 + 150Y, where X and Y are two separate variables denoting the levels of productivity of two products.
Ascertain the levels of productivity of the two products that optimises profits.
Solution
dπ = 100 – 8X
- Y
dX
dπ = -X – 16Y
+ 150
dY
For optimising profits we should the derivatives to zero.
100 – 8X – Y = 0 …..Equation (1)
150 – X – 16Y = 0 …..Equation (2)
To solve the equation, we can multiply the Equation (1) by -16, resulting the following
-1600 + 128X + 16Y = 0 …..Equation (3)
+ 150 - X - 16Y = 0 …..Equation (2)
-1450 + 127X = 0
127X = 1450
X = 1450 / 127
X = 11.42
Substituting the value of X = 11.42 in the Equation (1) we get the following.
100 – 8*11.42 – Y = 0
Y = 8.64
Therefore, the firm will maximise profits of its productivity and sells 11.42 units of products X and 8.64 units of product Y.
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- Basic Model of the Firm and Role of Profits
- Demand and Demand Function
- Macro Economics Policy
- Managerial Assessment Making Procedure
- Marginal and Incremental Analysis
- Maximisation by Marginal Examination
- Optimisation Concepts and Techniques
- Restrained Maximisation
- Restrained Maximisation: Substitution Method
