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The postal service will accept a box for shipment only if the sum of its length and girth (the distance around) does not exceed 108 inches. What dimensions will give a box with a square end the largest possible volume?

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    $\begingroup$ This is a pretty straightforward problem; have you at least made some progress towards setting it up? $\endgroup$ Dec 9, 2012 at 22:29
  • $\begingroup$ Could you also include a picture, labelled with appropriate variables? Doing so would show that you've put at least some effort into solving the problem... $\endgroup$
    – apnorton
    Dec 9, 2012 at 22:31
  • $\begingroup$ I know that you're trying to maximize volume which equals LWH. Since it's a square you know that W and H are equal. Since Girth plus length equals 108, that means that L+2W+2H=108. You could then solve for L which would equal 108/4w, since w=h. $\endgroup$ Dec 9, 2012 at 22:32
  • $\begingroup$ What you say will get you well down the path to solution. $\endgroup$ Dec 9, 2012 at 22:51

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Let $x$ be the length of a side of the square end, and let $y$ be the length of the box. As you say in your comment, the volume is given by $V=x^2y$, and the postal service’s restriction limits $4x+y$ to a maximum of $108$. Clearly we should use the entire allowance, so we want $4x+y=108$. However, you’ve solved this incorrectly for $y$: $y=108-4x$, not $\frac{108}{4x}$. Substituting the correct expression for $y$ into the volume formula, we get $V=x^2(108-4x)=108x^2-4x^3$.

You want to find the value of $x$ that maximizes $V$, knowing that $V=108x^2-4x^3$. How can you use the first derivative $\frac{dV}{dx}$ to find this value of $x$? (I’ll stop here to give you a chance to think about it; if you get completely stuck, leave a comment.)

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  • $\begingroup$ I guess it was just a careless error. I got 18 for height and width, and 36 for length. Thanks for your help. $\endgroup$ Dec 9, 2012 at 22:51
  • $\begingroup$ @Justin: You’re welcome. $\endgroup$ Dec 9, 2012 at 22:52
  • $\begingroup$ You also obviously need both $x,y$ non-negative, otherwise there is no maximum ($\lim_{x\to -\infty} V(x) = + \infty$). $\endgroup$
    – copper.hat
    Dec 9, 2012 at 23:06
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You have $L+4W = 108$, and you wish to maximize $L W^2$, subject to $L\geq 0, W \geq 0$. Since $L = 108-4W$, you want to maximize $f(W) = (108-4W)W^2 = 108W^2-4W^3$ subject to $W \geq 0$ and $L=108-4W \geq 0$, or $W \in [0, 27]$. We notice that $f(0) = f(27) = 0$ and $f(W) < 0$ for $W \in (27,\infty)$, hence the maximum occurs in $(0,27)$.

Setting $f'(W) = 0$ gives $W=0$ or $W = 18$, hence the solution must be $W=18$.

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You're pretty close with what you've got so far. Note that $108=L+2W+2H=L+4W$ implies that $L=108-4W$, not $L=\frac{108}{4W}$.

We're trying to maximize volume, so since $H=W$ and $L=108-4W$, then $$V=LWH=(108-4W)W^2=108W^2-4W^3.$$ Now, before we jump into the calculus side of things, let's see what values of $W$ even make sense. We can't have negative length, so we need $0\leq L=108-4W$, which implies that $W\leq 27$. On the other hand, we can't have negative width, either, so $0\leq W$, so the domain where this definition of volume in terms of $W$ makes sense would be $[0,27]$. If we check at the endpoints of the interval, we find $V(0)=V(27)=0$, so the maximizing width will be some $W$ strictly between $0$ and $27$. We'll find that by taking the derivative with respect to $W$, and setting it equal to $0$, giving us $$0=V'(W)=216W-12W^2=12W(18-W).$$ We'll need $W=18$, then. Knowing $W$, and knowing that $H=W$ and $L=108-4W$, we can the find the appropriate dimensions.

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