Acommunity 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?

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

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