Question

3. The profit function of a firm to produce two goods is given as T = 24Q1Q - Q, 0, - 203 +33Q, - 43 %3D Find the level of output required to maximize profit. Use the second degree differentiation to determine the maximum stagnation point.

Answer 1

Given

$T=24Q_1-Q_1^2-Q_1Q_2-2Q_2^2+33Q_2-43$

First, we will solve the firm's problem which is to maximize the profit of each good.

First-degree differentiation:

Differentiating profit function with respect to each good and equating to zero.

$\frac{dT}{dQ_1}=24-2Q_1-Q_2=0.....(1)$

$\frac{dT}{dQ_2}=-Q_1-4Q_2+33=0......(2)$

We have 2 variables and 2 equations,

$Equation_1:2Q_1+Q_2=24$

$Equation _2:Q_1+4Q_2=334$

Solving the equations:

Multiplying the equation 2 by 2 and subtracting from the equation 1.

We get,

9 units of good 1 and 6 units of good 2 will maximize the profit.

9 units of good 1 and 6 units of good 2 will maximize the profit.

 

Maximum profit:

$T=24Q_1-Q_1^2-Q_1Q_2-2Q_2^2+33Q_2-43.$

$T=24(9)-(9)^2-9\times 6-2(6)^2+33(6)-43=164$

Second-degree differentiation to check this is the maximum point.

If the second degree is positive, it is minimum and if it is negative, it is maximum.

$\frac{dT}{dQ_1}=24-2Q_1-Q_2$

$\frac{d^2T}{dQ_1^2}=-2$

$\frac{dT}{dQ_2}=-Q1-4Q_2+33$

$\frac{dT}{dQ_2}=-Q1-4Q_2+33$

$\frac{d^2T}{dQ_2^2}=-4$

Both are negative indicative negative stagnation points implying the maximum profits.

Stagnation points for Q1=-2 and for Q2=-4