solving 5000-50p^2 help me please!!!!

kimbero5

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Sep 11, 2005
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No. 10 (Market research)

The market research department of a company recommended to management that the company manufactures and markets a promising new product. After extensive surveys, the research department backed up the recommendation with the demand equation
x = 5,000 - 50p ,
Where x is the number of units that retailers are likely to buy per month at $p per unit.
Notice that as the price goes up, the number of units goes down.
From the financial department, the following cost equation was obtained:
C = 40,000 + 12x .

The revenue equation (the amount of money, R, received by the company for selling x units at $p per unit) is
R = xp , i.e. Revenue = (Number of units)(Price per unit).

(a) Express revenue R as a quadratic function of price p.
P(R)=5000p-50p ^2 right?

(b) Find the price that produces the maximum revenue.
for this one you set the problem = to zero and solve for it but i'm having trouble doing that it's been a while since i've done that...
 
(a) Express revenue R as a quadratic function of price p.
R(p)=5000p-50p<SUP>2</SUP> right? YES

(b) To find the price that produces the maximum revenue,
you need the derivative, marginal revenue, R’(p)=5000−100p.
Now solve R’(p)=0.
 
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