SOLUTION: This one was a bit tricky for me, If anyone can help, I will greatly appreciate it. Here goes: Demand for pools. Tropical Pools sells an aboveground model for p dollars each. The

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: This one was a bit tricky for me, If anyone can help, I will greatly appreciate it. Here goes: Demand for pools. Tropical Pools sells an aboveground model for p dollars each. The      Log On


   

Question 320043: This one was a bit tricky for me, If anyone can help, I will greatly appreciate it. Here goes:
Demand for pools. Tropical Pools sells an aboveground
model for p dollars each. The monthly revenue for this
model is given by the formula
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
to solve each problem.
Price (dollars)
Revenue (thousands
300
200
100
0
1000 2000 3000 4000
Price (dollars
Answer by ankor@dixie-net.com(22740) About Me  (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 for this model is given by the formula
R(p)= -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(p)= p(-0.08p + 300).
:
b) Write a formula D(p) for the monthly demand.
Demand is equal the revenue divided by the price
D(p) = p%28-.08p+%2B+300%29%2Fp
cancel p;
D(p) = -.08p + 300
:
c) Find D(3000).
D(3000) = -.08(3000) + 300
D(3000) = -240 + 300
D(3000) = 60 pools when the price is $3000
:
d) Use the accompanying graph to estimate the price at
which the revenue is maximized.
+graph%28+300%2C+200%2C+-1000%2C+5000%2C+-50000%2C+400000%2C+-.08x%5E2%2B300x%29+
$1900 price for max revenue
:
Approximately how many pools will be sold monthly at this price?
28000/1850 = 147 pools sold
:
e) What is the approximate maximum revenue?
$280,000
:
f) Use the accompanying graph to estimate the price at which the revenue is zero
$3800
: