Question #218139

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?

Expert's answer

A.

Profit=Revenue−CostProfit=Revenue-Cost

P(Q)=R(Q)−TC(Q)P(Q)=R(Q)-TC(Q)


R(Q)=p(Q)Q=(10−0.001Q)QR(Q)=p(Q)Q=(10-0.001Q)Q

=10Q−0.001Q2=10Q-0.001Q^2

P(Q)=10Q−0.001Q2−(5000+2Q)P(Q)=10Q-0.001Q^2-(5000+2Q)

=8Q−0.001Q2−5000=8Q-0.001Q^2-5000


P(Q)=8Q−0.001Q2−5000P(Q)=8Q-0.001Q^2-5000

B.


Q≥0,p(Q)≥0Q\geq0, p(Q)\geq0

10Q−0.001Q2≥010Q-0.001Q^2\geq0

0≤Q≤100000\leq Q\leq 10000

P′(Q)=(8Q−0.001Q2−5000)′P'(Q)=(8Q-0.001Q^2-5000)'

=8−0.002Q=8-0.002Q

Find critical number(s)


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

If 0≤Q≤4000,P′(Q)>0,P(Q)0\leq Q\leq4000, P'(Q)>0, P(Q) increases.


If 4000≤Q≤10000,P′(Q)<0,P(Q)4000\leq Q\leq10000, P'(Q)<0, P(Q) decreases.

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


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


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

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

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

p(4000)=$6p(4000)=\$6
LATEST TUTORIALS
APPROVED BY CLIENTS