Answer to Question #216770 in Calculus for sahal

Question

A company manufactures and sales x computers per month. The monthly cost and price demand equation are C(x) = 72000 + 60𝑥 Price-demand equation is 𝑝(𝑥) = 200 − 𝑥 30 a. Find the maximum revenue. b. Find the maximum profit and the production level that will realize the maximum profit.  c. The price the company should sell for each computer. 

Answer 1

a. Revenue is

$$R(x)=p(x)x=(200-\dfrac{x}{30})x=200x-\dfrac{x^2}{30}, x\geq0$$

$$R'(x)=(200x-\dfrac{x^2}{30})'=200-\dfrac{x}{15}$$

Critical number(s)

$$R'(x)=0=>200-\dfrac{x}{15}=0=>x=3000$$

If $0<x<3000, R'(x)>0, R(x)$ increases.

If $x>3000, R'(x)<0, R(x)$ decreases.

The function $R(x)$ has a local maximum at $x=3000.$

Since the function $R(x)$ has the only extremum for $x\geq 0,$ then the function $R(x)$ has the absolute maximum at $x=3000$ for $x\geq0.$

$$R(3000)=200(3000)-\dfrac{(3000)^2}{30}=\$300000$$

b.  Profit is

$$P(x)=R(x)-C(x)=200x-\dfrac{x^2}{30}-(72000+60x)$$

$$=200x-\dfrac{x^2}{30}-(72000+60x)$$

$$=120x-\dfrac{x^2}{30}-72000$$

$$P'(x)=(120x-\dfrac{x^2}{30}-72000)'=120-\dfrac{x}{15}$$

Critical number(s)

$$P'(x)=0=>120-\dfrac{x}{15}=0=>x=1800$$

If $0<x<1800, P'(x)>0, P(x)$ increases.

If $x>1800, P'(x)<0, P(x)$ decreases.

The function $P(x)$ has a local maximum at $x=1800.$

Since the function $P(x)$ has the only extremum for $x\geq 0,$ then the function $P(x)$ has the absolute maximum at $x=1800$ for $x\geq0.$

$$P(1800)=120(1800)-\dfrac{(1800)^2}{30}=\$108000$$

c.

$$p(1800)=200-\dfrac{1800}{30}=\$140$$