Answer to Question #218139 in Calculus for Alex

Question

An office supply company sells Q permanent markers per year at $P per marker. The price-demand equation for these markers is P=10-0.001Q. The total cost of manufacturing Q permanent markers is TC(Q)=5000+2Q.

A. What is the company's maximum profit?B. What should the company charge for the markers, and how many markers should be produce to maximize the profit?

Answer 1

A.

$$Profit=Revenue-Cost$$

$$P(Q)=R(Q)-TC(Q)$$

$$R(Q)=p(Q)Q=(10-0.001Q)Q$$

$$=10Q-0.001Q^2$$

$$P(Q)=10Q-0.001Q^2-(5000+2Q)$$

$$=8Q-0.001Q^2-5000$$

$$P(Q)=8Q-0.001Q^2-5000$$

B.

$$Q\geq0, p(Q)\geq0$$

$$10Q-0.001Q^2\geq0$$

$$0\leq Q\leq 10000$$

$$P'(Q)=(8Q-0.001Q^2-5000)'$$

$$=8-0.002Q$$

Find critical number(s)

$$P'(Q)=0=>8-0.002Q=0=>Q=4000$$

If $0\leq Q\leq4000, P'(Q)>0, P(Q)$ increases.

If $4000\leq Q\leq10000, P'(Q)<0, P(Q)$ decreases.

The function $P(Q)$ has a local maximum at $Q=4000.$

Since the function $P(Q)$ has the only extremum on $[0, 10000],$ then the function $P(Q)$ has the absolute maximum at $Q=4000.$

$$P(4000)=8(4000)-0.001(4000)^2-5000$$

$$P(4000)=\$11000$$

$$p(4000)=10-0.001(4000)$$

$$p(4000)=\$6$$