2
advanced-math

What is the maximum revenue for R(p)=−4p2+400pR(p) = -4p^2 + 400p?

A

$10,000

B

$20,000

C

$5,000

D

$40,000

Correct Answer: A

Choice A is the correct answer. Substitute the optimal price (p=50p=50) back into the revenue equation.

  1. R(50)=−4(50)2+400(50)R(50) = -4(50)^2 + 400(50).
  2. R(50)=−4(2500)+20000R(50) = -4(2500) + 20000.
  3. R(50)=−10000+20000=10000R(50) = -10000 + 20000 = 10000.

Choice B is incorrect. Choice C is incorrect. Choice D is incorrect.

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Set 18: Quadratic Equations and Functions 18