An office supply store sells about 8 graphing calculators per month for $120 each. (cont. in description)

Question

For each $6 decrease in price, the store expects to sell eight more calculators.The revenue from calculator sales is given by the function R(n) = (unit price)(units sold), or R(n) = (120 - 6n)(80 + 8n), where n is the number of $6 price decreases.

 

a. How much should the store charge to maximize monthly revenue?

 

b. Using a different revenue model, the store expects to sell 5 more calculators for each $4 decrease in price. Which revenue model results in great maximum monthly revenue? Explain.

Answer 1

a. I think that you mean:  at a price of $120, 80 calculators are sold.

 

    Then, R(n) = (120-6n)(80+8n)

 

    The graph is a parabola opening downward.  R(n) = 0 when n = 20 or n = -10

 

           By the symmetry of the parabola, the maximum occurs halfway between -10 and 20.

 

           So, maximum revenue when n = 5 (that is, decrease the price 5 times by $6 each time)

 

                 Maximum revenue occurs when the unit price is $90.

 

b.  Let n = number of $4 decreases

 

     Number of calculators sold = 80 + 5n

 

     Price per calculator = 120 - 4n

 

     R(n) = (price per calculator)(number sold) = (120 - 4n)(80 + 5n)