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October 1, 2014

October 1, 2014

Posted by **Nina** on Sunday, February 22, 2009 at 4:47pm.

R (p) = - 0.08p2 + 300p.

Revenue is the product of the price p and the demand (quantity sold).

a) Factor out the price on the right-hand side of the formula.

b) Write a formula D (p) for the monthly demand.

c) Find D (3000).

d) Use the accompanying graph to estimate the price at which the revenue is maximized. Approximately how many pools will be sold monthly at this price?

e) What is the approximate maximum revenue?

f) Use the accompanying graph to estimate the price at which the revenue is zero.

- Algebra 2 -
**Anonymous**, Monday, April 13, 2009 at 9:45pmR= - 0.08p^2 + 300p

Revenue is the product of the price P and the demand (quantity sold).

A) Factor out the price on the right-hand side of the formula.

R= - 0.08p^2 + 300p = p ( -0.08p + 300)

B) What is an expression for the monthly demand?

since revenue is price times demand, then the demand is:

-0.08p + 300

C) What is the monthly demand for this pool when the price is $3000?

p=3000

demand= -0.08(3000) + 300 = 60

D) Estimate the price at which the revenue is maximized. Approximately how many pools will be sold monthly at this price?

the maximum of the revenue occurs when R'= -2(0.08)p+300 =0

i.e when p= 300/2(0.08) = 1875.

E) what is the approximate maximum revenue?

p=1875

R=-0.08(1875)^2 + 300(1875) = 281 250.00.

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