SOLUTION: Can anyone please offer some direction??? THANKS SOOO MUCH!! Demand for pools. Tropical pools sells an above ground model for p dollars each. The monthly revenue from the sale of

Question 142889: Can anyone please offer some direction??? THANKS SOOO MUCH!!
Demand for pools.
Tropical pools sells an above ground model for p dollars each. The monthly revenue from the sale of this model is given by
R= -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.
b. What is an expression for the monthly demand?
c. What is the monthly demand for this pool when the price is $3,000.
d. Use the graph to estimate the price at which the revenue is maximized. Approximately how many pools will be sold monthly at this price?
(I can�t scan the graph, but the price at which the revenue is maximized is at approx. $2,000)
e. What is the approximate maximum revenue?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Demand for pools.
Tropical pools sells an above ground model for p dollars each. The monthly revenue from the sale of this model is given by
R= -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.
p(0.08p+300)
-----------------------
b. What is an expression for the monthly demand?
D(p) = -0.08p+300
----------------------
c. What is the monthly demand for this pool when the price is $3,000.
D(3000) = -0.08*3000+300 = 60 units
------------------------
d. Use the graph to estimate the price at which the revenue is maximized.

max occurs at p = -b/2a = -300/(2*-0.08) = $1875
------------------------
e. Approximately how many pools will be sold monthly at this price?
D(1875) = -0.08*1875+300 = 150 units
----------------------
e. What is the approximate maximum revenue?
R(p) = -0.08p^2 +300p
R(1875) = 0.08(1875)^2 + 300(1875) = $281,250
=====================
Cheers,
Stan H.

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