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