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.
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
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