Question

A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $6, and the society sells an average of 20 per week at a price of $10 each. The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it loses 2 sales per week.a) Find a function that models weekly profit in terms of price per feeder.b) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?

Answer 1

Answer:

A) (20-(x-10)) (x-6) = -x^2 +36x-180

B)In order to maximize the weekly profit the feeder price should be $80 and the weekly profit will be $144

Explanation

A)Let's consider they priced each feeder dollar x then the gain (price minus cost) of each feeder is (x-6). With the price of Dollar 10 they can sell 20 per week.

For every dollar increase in price they will lose two sales so total sale is

(10-(x-10)) .By multiplying the number of sales of feeder with profit of each feeder sold we get total revenue y i.e.

Y=(x-6)(20-(x-10))

=(x-6) (30-x)

= -x^2+36x-180

B) compare above equation with quadratic equation we get;

a=-1 , b=36 , c= -180

A quadratic function has minimum or maximum at point -b/2a ,so revenue is increased for feeder price.

x= b/2a = 36/2(-1) = $18

AND

Maximum weekly profit is

Y(18) = (18-6) (30-18)

=$144