Question

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

Answer 1

${\displaystyle C\left(x\right)=\frac{x^3}{3}-45x^2+900x+36}$
${\displaystyle \frac{dc\left(x\right)}{dx}=x^2-90x-900=M\left(x\right)}$

= Marginalcost
${\displaystyle \frac{dc\left(x\right)}{dx}=x^2-90x-900=M\left(x\right)}$
                                              = marginal cost
${\displaystyle \frac{d^{2}c\left(x\right)}{{dx}^2}=\frac{2x-90=dM\left(x\right)}{dx}}$
 
${\displaystyle \therefore \frac{d^{2}M\left(x\right)}{{dx}^2}=2>0}$

${\displaystyle \therefore \frac{dM\left(x\right)}{dx}}$ is minimum
To be minimum
${\displaystyle \frac{dM\left(x\right)}{dx}=0}$
2x - 90 =0
2x = 90
x= 45