What should be the selling price be in order to maximize revenue?

Question

A company is selling sweaters as a fundraiser. They sell 1200 a year at $20 each. They are planning to increase the price. However, a survey indicates that for every $2 increase in price, there will be a drop of 60 sales a year.

Answer 1

let x=# of increases in price and number of (60) decreases in sales

(1200-60x)(20+2x)=0

24,000-1200x+2400x-120x²=0

24,000+1200x-120x²=0

divide by 120

200+10x-x²=0

use -b/2a

-10/-2=5 $2.00 increases will maximize revenue

$20+5($2)=$30 per sweater and 1200-5(60)=1200-300=900 sales

900*$30=$27,000 revenue

 

# sold     price        revenue

1200        $20        $24,000

1140        $22        $25,080

1080        $24        $25,920

1020        $26        $26,520

960          $28        $26,880

900          $30        $27,000

840          $32        $26,880

780          $34        $26,520

Answer 2

Sales = Sweaters * Price

 

Initially, Sales = 1200*$20

 

Taking into account the survey results, you get:

 

S(x) = x[1200-(60/2)(x-20)] ~  This subtracts off the lost 60, per each $2 increase.  This is a quadratic (parabola), and it's vertex will be your maximum

 

S(x) = x[1200-30(x-20)]

S(x) = -30x² + 1800x

 

Taking the easy way out, and using the vertex formula of (-b/2a, f(-b/2a)), you get this vertex to be at x=30, which is your selling price.