# How to find Perimeter and Area?

I have this question: A rancher has 480 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be a maximum? x=....ft y=....ft

Below is my work:

1. P=4x+2y

2. A=2xy

480=4x+2y

240-2x=y

A=2x(240-2x)

480x-4x^2

I cannot figure out what do I have to do next??? Thanks in advance.

• What are $x$ and $y$? – Chris Eagle Jul 1 '13 at 15:41
• @Sara, I think $x,y$ are the length & breadth. But, could you draw the picture ? – lab bhattacharjee Jul 1 '13 at 15:44
• We could always make a big corral and a micro-corral stuck to the side of the big one, for the use of a micro-horse. It would have been nice if the problem had specified the meaning of adjacent, presumably that they share a full side. – André Nicolas Jul 1 '13 at 16:30

Because there are two adjacent rectangular corrals, we need for perimeter to be such that $$P = 3x+2y = 480\iff 2y = 480 - 3x \iff y = 240 - \frac 32 x$$

Where $3x$ counts the three sides, one of which is shared by each rectangle.

Then Area is $x y$...$y$ being the width across the large rectangle, and $x$ being the length of the large rectangle containing the adjacent rectangles.

$$A = xy = x\left(240 - \frac 32 x\right) = 240 x - \frac 32 x^2$$

Now find $A'(x)$ and set that equal to $0$ and solve for maximum $x$.

$$A'(x) = 240 - 3x = 0 \iff 3x = 240 \iff x = 80\;\text{feet}$$

$$y = 240 - \frac 32 x = 240 - 120 = 120\;\text{feet}$$

So the large rectangle will be 120 feet across top and bottom, and 80 feet along each side: Two outer sides, one inner (middle/dividing) side.

• Well that's what I gat x=80, and y=120 and it was wrong? – Sara Sharp Jul 1 '13 at 16:01
• Well, it's correct, perhaps a typo in the solutions? – Namaste Jul 1 '13 at 16:08
• Each of the two adjacent corrals will then by $x \times y/2:$ 80 feet long, 60 feet wide. – Namaste Jul 1 '13 at 16:11
• I came to this answer: x=60, y=240/3. I'm not sure if I'm correct? – Sara Sharp Jul 1 '13 at 16:37
• Yes, did you see my comment: we could have 80 by 60 rectangles, but there'd be two of them: two 60 feet by 80 feet, but they'd be side by side, sharing an 80 foot fence. So we still get the big rectangle 120 feet by 80 feet. It depends on what you call x, y: in essence, we'd have three sides of fencing each 80 feet long, and 4 sides of fencing, each 60 feet. Two of each of the 60 ft fencing will be joined to make a long two 120 ft long fences. Just as long as you make clear what $x$ and $y$ each mean, you're fine with the answer. – Namaste Jul 1 '13 at 16:44