Hi Kay :))
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.
At this point the feeders will cost 6*20 = $120
And will sell for 10*20= $200
So profit will be 10*20 - 6*20 = $80
The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it will lose 2 sales per week.
(a) Find a function that models weekly profit in terms of price per feeder.
Let d be the number of extra dollars that will be charged.
The number that will be made and sold will be N=(20-2d) The sale price will be (10+d)
Total cost will be 6(20-2d) dollars
The total revenue (sale) will be (10+d)(20-2d)
so
Profit = revenue - cost
Profit = (10+d)(20-2d) - 6(20-2d)
Profit = (10+d-6)(20-2d)
Profit = (d+4)(20-2d)
Profit = (d+4)(20-2d)
I just realized that I am meant to be getting profit in relation to Price per feeded so I will substitute for d
Let x be the sales price.
\(x=10+d\\ d=x-10\\\)
\(Pr=(d+4)(20-2d)\\ Pr=(x-10+4)(20-2(x-10))\\ Pr=(x-6)(20-2x+20)\\ Pr=(x-6)(40-2x)\\ Pr=-2(x-6)(x-20)\\ \)
(b) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?
Profit as a funtion of sales price is a concave down parabola.
The roots (where there is no profit) are at Price=$6 and $20
The maximum profit will be half way between these 2 where x=(6+20)/2= $13
If $13 is charged it will yeild the maximum profit of
\(Pr=-2(13-6)(13-20)\\ Pr=-2*7*-7\\ Pr=$98\)
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