Question

a manufacture can sell x times. determine the total revenue function,profit function and level of output that will maximize profit

Answer 1

A manufacturer can sell x items at the price of Rs. (330−x) each. The cost of producing x items is Rs. $x^2+10x-12$

The total revenue function

$$R(x)=xp(x)=x(330-x)=330x-x^2$$

The cost function

$$C(x)=x^2+10x-12$$

Profit

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

$$P(x)=330x-x^2- (x^2+10x-12)=-2x^2 +320x+12$$

We need maximize the profit.

Find the first derivative with respect to x

$$P'(x)=(-2x^2+320x+12)'=-4x+320$$

Find the critical number(s)

$$P'(x)=0=>-4x+320=0=>x=80$$

First derivative test

If $x<80,$ then $P'(x)>0, P(x)$ increases.

If $x>80,$ then $P'(x)<0, P(x)$ decreases.

$$P(80)=-2(80)^2+320(80)+12=12812$$

The profit function $P(x)$ has the local maximum with value of $12812$ at $x=80.$

Since $x=80$ is the only critical number, then the profit function $P(x)$ has the absolute maximum with value of $12812$

at $x=80.$

The manufacturer can earn the maximum profit with value of $Rs. 12812$ , if he sells $80$ items.