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The price p in dollars and the quantity x sold of a certain product obey the demand equation -1/5x+100 (0≤ x ≤ 500)
a) Express the revenue R as a function of x. (remember, R= p.x)

b) What is the revenue if 200 units are sold?

c) What quantity x that maximizes the revenue?

e) What price should the company charge to maximize revenue?

Respuesta :

mhanifa

Answer:

  • See below

Step-by-step explanation:

We have:

  • p = -1/5x + 100 = -0.2x + 100

a) The revenue is:

  • R = px
  • R = x( -0.2x + 100)
  • R= -0.2x² + 100x

b) x = 200, find R:

  • R = -0.2(200²) + 100(200) = 12000

c) This is a quadratic function and the maximum value is obtained at vertex.

  • x = -100/(-0.2*2) = 250 is the required quantity

d) The max revenue is obtained when -0.2x + 100 at max:

  • The maximum possible is p = 100 when x = 0

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