Hi there!
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.
R = p(-0.08p + 300)
b) What is an expression for the monthly demand?
D = R/p
D = -0.08p + 300
c) What is the monthly demand for this pool when the price is $3000?
D = -0.08*3000 + 300
D = 60
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?
Maximum is at the vertex...
-b/2a
= -300/(2*-0.08)
Revenue is maximized when P = 1875 (you should see this on the graph that you have)
When P = 1875:
D = -0.08*P + 300
D = 150
e) What is the approximate maximum revenue?
The maximum revenue is:
-0.08*1875^2 + 300*1875
= 281250
Let me know if you have any questions. If not, thanks for pressing "Accept".
-Scott
MIT Graduate
College degree in math... proficient in all levels -- from algebra to calculus