# Math Help - optimization problem

1. ## optimization problem

A farmer wishes to fence an area next to his barn. He needs a wire fence that costs $1 per linear foot in front of the barn and and wooden fencing that costs$2 per foot on the other sides.
Find x and y so that he can enclose the maximum area if his budget for materials is $4400. $A=2xy$ $c=(2*2x) + (1*y)=4400$ $4x=4400-y$ $x=1100-\frac{y}{4}$ $A=2(1100-\frac{y}{4})y=2200y-\frac{y^2}{2}$ $A'=2200-y$ $y=2200 $ Y is supposed to be $\frac{2200}{3}$. So either my area function or my constraint is wrong? 2. Does the problem say rectangular area? If so, You're area should just be xy, if you draw a picture you will see this, you need the 2 in there when considering cost, as you have done 3. why do you think Y is incorrect? because I am actually getting 2200 and 550 4. They had a picture like this: where W was labled " $x$", and L was labeled " $y$". The top horizontal piece opposite the "L" was labeled " $barn(wire)$." I still ended up getting $y=2200$, when I did $A=xy$though. $A=(1100-\frac{y}{4})y$ $1100y-\frac{y^2}{4})$ $A'=1100-\frac{y}{2})$ $1100=\frac{y}{2}$ $y=2200$ ? 5. Originally Posted by artvandalay11 why do you think Y is incorrect? because I am actually getting 2200 and 550 Here are the solutions: http://www.math.ufl.edu/~jysmith/MAC...0answerf09.pdf This is #22. 6. Yes I'm sorry about the xy comment, I realized it wouldnt make a difference for your Y value since the 2 goes away when you differentiate, but that answer has to be incorrect We both agree on Cost: $C=2(2x)+1(y)=4x+y=4400$ Now let's try the "answer" $4(550)+\frac{2200}{3}=2200+\frac{2200}{3}\not =4400$ 7. Y would equal $\frac{2200}{3}$ if the cost for that side of fencing was$3 per foot

8. That is true. I guess I'll have to straighten that out with my professor.Well, anyways .