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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?

enter image description here

*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.

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    $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Aug 13, 2020 at 23:48
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    $\begingroup$ post edit: happily the picture indicates which dimension is the width, and thereby removes the potential ambiguity. $\endgroup$
    – lulu
    Commented Aug 14, 2020 at 0:00
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    $\begingroup$ 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. $\endgroup$
    – Doug M
    Commented Aug 14, 2020 at 0:08

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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.

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