Question

The total cost function of a firm is $C=x^2+75x+1600$ for output $x$. Find the output $\left(x\right)$ for which average cost is minimum. Is $C_A=C_M$ at this output?

Answer 1

We have,
$C\left(x\right)=x^2+75x+1600$
$\overline{C}\left(x\right)=\frac{C\left(x\right)}{x}$
$\overline{C}\left(x\right)=\frac{x^2+75x+1600}{x}$
$\overline{C}\left(x\right)=x+75+\frac{1600}{x}$
Now,
$\overline{C^{\prime }}\left(x\right)=\frac{d\overline{C}\left(x\right)}{dx}$
$=1-\frac{1600}{x^2}$
For the minimum average cost $C^{\prime }\left(x\right)=0$
$1-\frac{1600}{x^2}=0$
$x=40$
Now,
$\overline{C"}\left(x\right)=\frac{d^{2}C\left(x\right)}{d{x}^2}$
So, $\frac{3200}{x^3}>0$ [for $x=40$]
Therefore, it is minimum.
Therefore, the minimum average cost
$\overline{C}\left(x\right)=40+75+\frac{1600}{40}=155$
Therefore, $C_A=155$
Now, we find marginal cost i.e.,
$C_m=\frac{dC}{dx}$
$C_m=\frac{d}{dx}(x^2+75x+1600)$
$C_m=2x+75$
Put $x=40$, we get
$C-m=2\times 40+75$
$C_m=80+75=155$
Thus, $C_A=C_m$ for $x=40$
Hence, this is the answer.