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I am asked to find the maximal volume of a rectangular box with a fixed surface area of $150$. I would like to solve this problem using gradients/Lagrange multipliers. I have done a bit of work so far but I'm not sure if I am on the right track or how to complete the problem. I know that this box should be a cube, but I would like to show this without assuming it is true. Here is what I have so far:

$Volume=V(l,w,h)=lwh$

$Surface Area=S(l,w,h)=2(lw+hw+hl)$.

$\triangledown V(l,w,h)= \lambda \triangledown S(l,w,h)$

$\triangledown V(l,w,h)=(wh)\hat{i} + (lh)\hat{j} + (wl)\hat{k}$

$\triangledown S(l,w,h)=(2w+2h)\hat{i} + (2l+2h)\hat{j} + (2w+2l)\hat{k}$

So putting this all together gives me:

$(wh)\hat{i} + (lh)\hat{j} + (wl)\hat{k} =\lambda [(2w+2h)\hat{i} + (2l+2h)\hat{j} + (2w+2l)\hat{k}]$

And at this point I am not sure how to conclude that $w=h=l$ .

Additionally, I see that I can look at the system of equations:

$wh=\lambda2(w+h)$

$lh=\lambda2(l+h)$

$wl=\lambda2(w+l)$

$2(lh+wh+lw)=150$

But I am unsure of how to solve these.

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    Any reason why you *want* to solve it using Lagrange multipliers? This is a classic AM-GM problem. – 2017-02-08
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    You're almost home. From $$(wh)\hat{i} + (lh)\hat{j} + (wl)\hat{k} =\lambda [(2w+2h)\hat{i} + (2l+2h)\hat{j} + (2w+2l)\hat{k}]$$, you have $wh = 2(w+h)$, therefore $V/\lambda = whl/\lambda = 2(wl+hl)$, similarly, you can derive $whl/\lambda = 2(lw+hw)$, these two lead to $hl = hw$, $l = w$. – 2017-02-08
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    @dxiv We are working on this technique in class right now but I can look into that as well. – 2017-02-08
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    @Guangliang Thanks, that makes perfect sense! – 2017-02-08
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    Once I see that $h=w=l$, I find that $S(l, w, h)=2(lw+hw+lh)=2(3w^2)$, $6w^2=150$, $w=h=l=5$ and so the maximal volume of this box equals $5^3=125$ – 2017-02-08

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Hint (without Lagrange multipliers): by AM-GM $\;\frac{1}{6}S=\frac{1}{3}(lw+hw+hl) \ge \sqrt[3]{l^2w^2h^2}=\sqrt[3]{V^2}\;$ with equality iff $lw=wh=hl \iff l=w=h\,$.