# Word Problem: Dimensions of the corral

Question: A two-pen corral is to be built. The outline off the corral forms two identical adjoining rectangles, as shown in the diagram below. If there is $$120$$m of fencing available and the fence width cannot be less than $$6$$m, what dimensions of the corral will maximize the enclosed area?

*I got really confused on this question as I don't know how to solve it. I am confused on what the question is asking me and what formula to use. I'd appreciate if anyone can help me solve this question.

• Well, I'd start by drawing a picture. Then label the relevant dimensions in your picture, express the constraint that you were given and then optimize the area subject to the constraint(s). Note: It's not at all clear to me from the question which dimension in the width, and that distinction might matter.
– lulu
Commented Aug 13, 2020 at 23:48
• post edit: happily the picture indicates which dimension is the width, and thereby removes the potential ambiguity.
– lulu
Commented Aug 14, 2020 at 0:00
• You can do this without calculus. $L = \frac {(120 - 3W)}{2}, A = LW = 60 W - \frac 32 W^2.$ Find the vertex of the parabola. Commented Aug 14, 2020 at 0:08

## 1 Answer

Define $$x$$ as the length of one pen of the corral and $$y$$ the width of the corral. Then you know that:

$$4x+3y = k, \:\: y \geq 6 , \:\: k \leq 120$$

And you want to find $$\max\{f(x)=Area(corral)=2xy\}$$ , so you can rewrite it as $$\max\{f(x) =Area(corral)=\frac{2x(k-4x)}{3}\}$$, then you impose the condition $$f'(x) = 0$$ to find the maximum which results in :

$$\frac{2k}{3}-\frac{16}{3}x=0 \iff x= \frac{k}{8}$$

So $$y=\frac{k}{6}$$ . Then , $$\max\{f(x)\} = k^2/24$$ , so the choice $$k=120$$ is the best one, then: $$x=15,\:y=20$$

I wanted to be formal defining $$k \leq 120$$ instead of putting just $$120$$ from the start but it's obvious that $$120$$ is the right one because the area of the corral is rectangular, so the more you increment the width or the length the more you cover area.