A carpentry shop currently sells 66000 decoy ducks per year at a cost of 16 dollars each. In the previous year when they raised the price to 21 dollars, they only sold 51000 decoy ducks that year. Assuming the amount of decoy ducks sold is in a linear relationship with the cost, what price should the carpentry shop charge per decoy duck to maximize their revenue?

Answer:19 dollarsStep-by-step explanation:Buckle up! This is a good one!First we have to find the line that goes trough the two points. And it has the form :[tex]y=mx+b[/tex]m is often called the slope.Given the two points (66000, 16) and (51000, 21) we find m with this formula:[tex]m=\frac{y_{2} -y_{1} }{x_{2} -x_{1} }\\m=\frac{21-16}{51000-66000} \\m=-\frac{5}{15000} \\m=-\frac{1}{3000}[/tex]We pick the point (51000,21) and plug it into the equation of the line and we find b:[tex]y= mx + b\\21=(-\frac{1}{3000} )(51000)+b\\21=-17+b\\21+17=b=38[/tex]And we have got our line equation[tex]y= -\frac{1}{3000}x + 38\\[/tex]Where x represents the quantity and y represents the price.The revenueThe revenue is price times quantity, since price is y and quantity is x, we get[tex]y=-\frac{1}{3000} x+38\\y*x=x*(-\frac{1}{3000} x+38)\\yx=-\frac{1}{3000} x^2+38x\\R(x)=-\frac{1}{3000} x^2+38x[/tex]to maximize the revenue, we get the first derivative:[tex]R(x)=-\frac{1}{3000} x^2+38x\\R'(x)=-(2)\frac{1}{3000} x+38\\R'(x)=-\frac{2}{3000} x+38\\\\R'(x)=-\frac{1}{1500} x+38\\\\[/tex]and we make R'(x)=0:[tex]R'(x)=-\frac{1}{1500} x+38=0\\38=\frac{1}{1500} x\\38*1500=x=57000[/tex]So the quantity that maximizes revenue is 57000, let's find out the price by plugging it into the equation:[tex]y= -\frac{1}{3000}x + 38\\y= -\frac{1}{3000}(57000) + 38\\y= -19 + 38\\y=19[/tex]And there you go! 19 dollars is the price the carpentry shop should charge per decoy duck.