SOLUTION: An ice cream company finds that at a price of $3.00 demand is 3500 units. For every $0.10 decrease in price, demand increases by 50 units. Find the price and quantity sold that max

Algebra ->  Equations -> SOLUTION: An ice cream company finds that at a price of $3.00 demand is 3500 units. For every $0.10 decrease in price, demand increases by 50 units. Find the price and quantity sold that max      Log On


   

Question 1104103: An ice cream company finds that at a price of $3.00 demand is 3500 units. For every $0.10 decrease in price, demand increases by 50 units. Find the price and quantity sold that maximizes revenue.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You can make a data table to help find the price and demand, and form your revenue function; price%2Aunits=revenue.

Let n be how many increments of $0.10.
PRICE per UNIT     Demand in UNITS
   3                        3500
   3-0.1                  3500+50
   3-2*(0.1)              3500+2(50)
   .
   .
   3-0.1n                3500+50n

highlight%28R%28n%29=%283500%2B50n%29%283-0.1n%29%29

You can leave R in its factored form. R is a parabolic function and has a maximum point. Find the ROOTS for R, and the maximum will happen exactly in the middle of the two values of n.

%283500%2B50n%29%283-0.1n%29=0-------find the two n values and identify the value exactly in the middle.

Do that and then finish the rest.