# Minimization word problem. Fence enclosure.

A farmer wishes to fence a rectangular pasture of area $2800~m^2$. One side of the fence is along a road, and that side costs \$5 per meter. The other 3 sides cost \$2 per meter.

Find the minimum possible cost.

I would be able to do this if all of the sides cost the same, but the fact that the one side costs \$5 throws me off. I understand that the formula will be$xy = 2800$. But I do not know how to represent the variables and isolate them. Any help is much appreciated! • Dollar signs carry special significance on this site (see this page for more details). Therefore, if you want actual dollar signs, you have to tell the site that those are actual dollar signs, and you do that by putting a backslash in front of them, like so: \$. – Arthur Aug 3 '17 at 6:31

Let $x$ be the length of the side along the road and let $y$ be the length of the sides perpendicular to the road. Then you know that $xy=2\,800$, which means that $y=\frac{2\,800}x$. He will pay $x\times5\$+x\times2\$+y\times2\$+y\times2\$$for the fence; in other words, the cost (in dollars) will be 7x+4y. But since y=\frac{2\,800}x, this means that the farmer will pay 7x+\frac{11\,200}x dollars. Let c(x)=7x+\frac{11\,200}x. Then c'(x)=7-\frac{11\,200}{x^2} and the only positive x such that c'(x)=0 is x=40. Furthermore, c'(x)>0 if c>40 and c'(x)<0 if x<40. Therefore, the minimum of c is c(40)=560 and the farmer will have to pay 560\$$.

The sides of the rectangular pasture are $x$ (meter) and $y$ (meter), where the side along the road has length $x$ (meter). The costs are therefore (in dollars):

$c(x,y)=5x+2x+2y+2y=7x+4y$

You have to minimize the function $c(x,y)=7x+4y$ with the constraint $xy=2800$.

Hence $y=\frac{2800}{x}$.

Thus your problem ist to minimize the function

$f(x)=7x+4 \frac{2800}{x}$.

• I think the cost function $c(x,y)$ should be $5x+2x+2y+2y$ – shwetha Aug 3 '17 at 6:43
• OOOps ! you are right ! – Fred Aug 3 '17 at 7:32