Solve the problem.
-The price p and the quantity x sold of a certain product obey the demand equation $$p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200$$ What quantity x maximizes revenue? What is the maximum revenue?
A) 600; $60,000
B) 900; $45,000
C) 300; $45,000
D) 1,200; $60,000

Question

Quizplus · Accepted Answer

Revenue is maximized at x = 600 with a maximum revenue of $60,000. Revenue, R, is given by R = px. Substituting the demand equation into the revenue formula gives R = (-1/6)x^2 + 200x. Differentiating with respect to x and setting the derivative equal to zero gives the quantity that maximizes revenue. Solving this gives x = 600. Substituting x = 600 into the revenue equation gives the maximum revenue of $60,000.

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Solve the Problem $p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200$