Question

A firm has the cost function $C=\frac{x^3}{3}-7x^2+111x+50$ and demand function $x=100-p$.
Find the profit maximising level of output $x$

Answer 1

Cost function is $C\left(x\right)=\frac{x^3}{3}-7x^2+111x+50$
Demand function is $x=100-p\Rightarrow p=100-x$
Revenue function, $R\left(x\right)=p.x=x(100-x)=100x-x^2...\left(1\right)$

Profit function, $P\left(x\right)=Revenue-Cost$
$=R\left(x\right)-C\left(x\right)$
$=100x-x^2-\frac{x^3}{3}+7x^2-111x-50$
$=-\frac{x^3}{3}+6x^2-11x-50...\left(2\right)$

Differentiating of equation $\left(2\right)$ w.r. to $x$, we get
$\frac{dP}{dx}=-x^2+12x-11....\left(3\right)$
Now, $\frac{dP}{dx}=0\Rightarrow -x^2+12x-11=0\Rightarrow x^2-12x+11=0$
$\Rightarrow x^2-11x-x+11=0$
$\Rightarrow x(x-11)-1(x-11)=0$
$\Rightarrow (x-1)(x-11)=0\Rightarrow x=1,11$

Again differentiating, we get
$\frac{d^{2}P}{d{x}^2}=12-2x....\left(4\right)$
at $x=1,\frac{d^{2}P}{d{x}^2}=10\Rightarrow \frac{d^{2}P}{d{x}^2}>0$ (Minimum value)
at $x=11,\frac{d^{2}P}{d{x}^2}=-10\Rightarrow \frac{d^{2}P}{d{x}^2}<0$ (Maximum value)

Hence, the profit maximising level of output $x=11$.