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.
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 $lw=wh=hl \iff l=w=h\,$.