Sarah is selling boxes of homemade cupcakes. She figures she can sell 200 boxes if she sells them for $8 each. for every fifty cents she raises the price,she figures that she'll sell 10 fewer boxes.

Question

1. What price should Sarah sell the boxes for to maximize her revenue? What will that revenue be?
2. Sarah decides that she doesn't really care about profit, she just wants as many people as possible to experience her cupcakes. However, she needs to make a revenue of $640 to break even. What price should Sarah sell the boxes at and how many boxes will she be able to sell?

Answer 1

(200,8) is one point on the graph with dollars on the y axis and boxes on the x axis

slope = -.5/10 =-5/100=-1/20

P=-x/20 + b

8 =-200/20 + b

b=8+10 = 18

at $18, no one buys any boxes

equation for price is

P=-x/20 + 18

Revenue = R = Px = -x^2/20 + 18x

R' = -x/10 + 18 =0

x/10 = 18

x= 180 boxes to maximize revenue

P=-180/20+ 18 = 18-9 = $9 = revenue maximizing price

R= 640 = -x^2/20 + 18x

x^2/20 -18x+640 =0

x^2 -360x -+12800=0

x= 360/2 + or - (1/2)sqr(360^2 -4(12,800)) = 180+or - 140 = 40 or 320 boxes to break even

P=--(320)/20 + 18 = 18-16 = $2 per box to break even at 320 boxes