If $C(x)=\frac{1}{3}x^3+x^2-15x+3000$ is the cost function then the minimum value of marginal cost function is :

-16

The correct answer is Option (3) → -16

The Marginal Cost function is,

$MC(x)=\frac{d}{dx}C(x)=\frac{d}{dx}\left(\frac{1}{3}x^3+x^2-15x+3000\right)$

$=x^2+2x-15$

for critical point,

$f'(c)=0$

$⇒2x+2=0$

$⇒x=-1$

$∴MC(-1)=(-1)^2+2(-1)-15$

$=-16$ → Minimum Value