Question

Cost of assembling $x$ wallclocks is $(\frac{x^3}{3}-40x^2)$ and labour charges are $500x$. Find the number of wallclocks to be manufactured for which marginal cost is minimum.

Answer 1

Cost of assembling$=(\frac{x^3}{3}-40x^2)$

Cost of labor$=500x$

$\therefore$Total cost function $=(\frac{x^3}{3}-40x^2+500x)=C\left(x\right)$

Marginal cost$=\frac{d}{dx}C\left(x\right)=\frac{d}{dx}\frac{x^3}{3}-40x^2+500x$

$=x^2-80x+500=MC$

Marginal cost is minimum when $\frac{dMC}{dx}=0$ &

$\frac{d^{2}MC}{d{x}^2}>0$

$\therefore \frac{d}{dx}MC=\frac{d}{dx}(x^2-80x+500)=0$

Or $2x-80=0$

Or $x=40$

$\frac{d^{2}MC}{d{x}^2}=2$, which is always $>0$

$\therefore$Minimum marginal cost at $x=40$