Answer to Question #218139 in Calculus for Alex

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?

1
Expert's answer
2021-07-19T05:48:30-0400

A.

Profit=RevenueCostProfit=Revenue-Cost

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


R(Q)=p(Q)Q=(100.001Q)QR(Q)=p(Q)Q=(10-0.001Q)Q

=10Q0.001Q2=10Q-0.001Q^2

P(Q)=10Q0.001Q2(5000+2Q)P(Q)=10Q-0.001Q^2-(5000+2Q)

=8Q0.001Q25000=8Q-0.001Q^2-5000


P(Q)=8Q0.001Q25000P(Q)=8Q-0.001Q^2-5000

B.


Q0,p(Q)0Q\geq0, p(Q)\geq0

10Q0.001Q2010Q-0.001Q^2\geq0

0Q100000\leq Q\leq 10000

P(Q)=(8Q0.001Q25000)P'(Q)=(8Q-0.001Q^2-5000)'

=80.002Q=8-0.002Q

Find critical number(s)


P(Q)=0=>80.002Q=0=>Q=4000P'(Q)=0=>8-0.002Q=0=>Q=4000

If 0Q4000,P(Q)>0,P(Q)0\leq Q\leq4000, P'(Q)>0, P(Q) increases.


If 4000Q10000,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)25000P(4000)=8(4000)-0.001(4000)^2-5000

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

p(4000)=100.001(4000)p(4000)=10-0.001(4000)

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

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