The cost function of a product is given by $C(x)=\frac{x^3}{3}-45x^2-900x + 36$, where x is the number of units produced. In order to minimise the marginal cost, how many units of the product must be produced?

45

The correct answer is Option (3) → 45

Given the cost function $C(x)=\frac{x^3}{3}-45x^2-900x + 36$

∴ Marginal cost = $MC=\frac{d}{dx}(C(x))$

$⇒MC = \frac{1}{3}.3x^2-45.2x-900.1 + 0$

$⇒MC = x^2 - 90x - 900$.

To find value(s) of x so that MC is minimum, we should find value(s) of x where $\frac{d}{dx}(MC) = 0$ and $\frac{d^2}{dx^2}(MC)>0$

Now $\frac{d}{dx}(MC)=2x-90$ and $\frac{d^2}{dx^2}(MC)=2$

$\frac{d}{dx}(MC)=0⇒2x-90=0⇒x=45$

When $x=45,\frac{d^2}{dx^2}(MC)=2>0$

Hence, marginal cost (MC) is minimum when $x = 45$.