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Solve the Problem p=16x+200,0x1,200p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200

Question 169

Multiple Choice

Solve the problem.
-The price p and the quantity x sold of a certain product obey the demand equation p=16x+200,0x1,200p = - \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

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