Question

Consider a firm is having monopoly in production of two goods \(X\) and \(Y\). The markets for the two goods do not interact. The profit function is given by

Hint:

For maximizing a two-variable profit function, set both first partial derivatives equal to zero and solve the resulting simultaneous equations.

Answer 1

Correct Answer: 142.5

Step 1: Write the profit function.
\[ \pi(x,y)=25x+13y-2.5x^2-y^2+xy-10 \]
To maximize profit, we find the first-order conditions with respect to \(x\) and \(y\).

Step 2: Differentiate with respect to \(x\).
\[ \frac{\partial \pi}{\partial x}=25-5x+y \]
For maximum profit,
\[ 25-5x+y=0 \] \[ y=5x-25 \]

Step 3: Differentiate with respect to \(y\).
\[ \frac{\partial \pi}{\partial y}=13-2y+x \]
For maximum profit,
\[ 13-2y+x=0 \] \[ x=2y-13 \]

Step 4: Solve the two equations.
Using
\[ y=5x-25 \] and
\[ x=2y-13 \]
Substitute \(y=5x-25\) into \(x=2y-13\):
\[ x=2(5x-25)-13 \] \[ x=10x-50-13 \] \[ x=10x-63 \] \[ 9x=63 \] \[ x=7 \]
Now substitute \(x=7\) in \(y=5x-25\):
\[ y=5(7)-25 \] \[ y=35-25 \] \[ y=10 \]

Step 5: Calculate maximum profit.
Substitute \(x=7\) and \(y=10\) in the profit function:
\[ \pi(7,10)=25(7)+13(10)-2.5(7)^2-(10)^2+(7)(10)-10 \] \[ =175+130-2.5(49)-100+70-10 \] \[ =175+130-122.5-100+70-10 \] \[ =142.5 \]

Step 6: Final conclusion.
Hence, the maximum profit is
\[ \boxed{142.5} \]