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 they sell an average of 20 per week at a price of $10 each. They have been considering raising the price, so they conduct a survey and find that for every dollar increase they lose 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? c) What is the maximum profit?

Answer:N(x) = 40 - 2xP(x) = -2x² + 52 x - 240maximum profit = 13Step-by-step explanation:given data feeder cost = $6average sell = 20 per weekprice = $10 eachsolutionwe consider here price per feeder = x and profit per feeder  id here formula   = x - 6 so that heretotal profit will beP (x)  = ( x - 6 ) Nxhere N(x) is number of feeders sold at price =  xso formula for N (x)  is here N(x) = 20 - 2 ( x - 10 )    N(x) = 40 - 2xso that P(x) = (x-6) ( 40 - 2x) P(x) = -2x² + 52 x - 240since here a = -2b = 52c = -240 a < 0so quadratic function have maximum value of c - [tex]\frac{b^2}{4a}[/tex] so it will be maximum value = -240 - [tex]\frac{52^2}{4(-2)}[/tex] maximum value = 98so here maximum profit attained at x = [tex]\frac{-b}{2a}[/tex]x = [tex]\frac{-52}{2(-2)}[/tex]x = 13maximum profit = 13