How much should Oscar charge per cap?

Question

Oscar sells souvenir caps to tourists in a beach town. He observed that if he charged $10 per cap, then he sold 32 caps per day. When he raised the price to $12.50 per cap, then he sold 28 caps per day. Using this data, the demand equation is . The equation for revenue, R, is .

How much should Oscar charge per cap?

To support your decision mathematically:

a) Find a function that models weekly revenue in terms of price per cap, p.

b) What price should Oscar charge so that his revenue is as large as possible? Be sure to clearly indicate how you arrived at that answer. Show your work! (*Even if using a calculator or graph, show how and why you took the steps you did, not just what buttons you pushed.) Write your statement in a complete sentence and provide a clear answer to the problem.

c) Is maximum revenue the same as maximum profit? Oscar wonders if the price that generates the maximum revenue is the same price that generates the maximum profit. Profit = Revenue - Cost. The materials for each cap cost $4.50. Use your model to answer Oscar's question. Explain your results in the context of the problem and be sure to show how you used your model to get those results.

Answer 1

(32,10), (28,12.5) are 2 points on his revenue graph

equation is -2.5/4 = -.625 =-5/8= (y-10)/(x-32)

8y-80=-5x+160

y= (-5x+240)/8= -5x/8 +30

P(c)=-5c/8+30

y intercept =(0,30)

x intercept =(48,0)

midpoint= (48/2,30/2)=(24,15)

$15 is the rev maximizing price with 24 caps sold

max revenue= 24x15=$252

midpoint creates max area of rectangle with the axes

or use calculus

R=cP=-5c^2/8+30c

R'=-5c/4+30=0

5c/4=30

c= (4/5)30 =24

P=-5c/8+30 = -5(24)/8+30 = $15

Profit maximizing price is different from the Revenue maximing price

C(c)=4.5c

R-C=-5c^2/8+25.5c

R'-C'=-5c/4+25.5=0

c=(4/5)25.5=20.4 caps

P=-5c/8+25.5=-5(5.1)/2+25.5=25.5/2=$12.75= profit maximizing price

max profit=(12.75-4.5)(20.4)=8.25(20.4)= 165+ 33= $198<max rev