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D=0.25(360-0.01p^(3))^(2) unit price that maximizes revenue

User Katana
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1 Answer

4 votes

Answer:

410,884 units and price is 17.26 for which revenue function is maximum

Step-by-step explanation:

Given function

D= 0.25 (360−0.01p^3)^2

Revenue function

R(p) = p(0.25(360 − 0.01p^3)^2)

R′(p) = (0.25(360 −0.01p^3) ^2) + p(.5(360−0.01p^3) (−0.03p^2)) R′(p)=0

(p, f(p)) = (17.2610874799436, 410,884.335440944)

(p, f(p)) = (33.0192724889463,0)

R″ = 0.25 (0.0042p^5 − 86.4p^2)

R″ < 0, at p = 17.26

410,884 units and price is 17.26 for which revenue function is maximum

User Tom Baxter
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