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
Algebra ->
Polynomials-and-rational-expressions
-> 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
Log On
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?
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.
(