A bulldozer manufacturer  produces x items and the total cost C and total revenue R are given by $C= 100 +0.015x^2$  and R =3x. The number of bulldozers to be produced for maximum profit is :

100

The correct answer is Option (3) → 100

Profit (P) is,

$P(x)=R(x)-C(x)$

$=3x-(100+0.015x^2)$

$=-0.015x^2+3x-100$

$\frac{dP}{dx}=-0.03x+3$

$=-0.03x+3=0$

$=x=100$ → Maxima occurs.