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I have this question, I have run completely blind into.

Find by Lagrange multipliers the volume V=xyz of where the largest box with sides adding up to x+y+z = k.

I have found the gradient of V:

dV/dx = yz
dV/dy = xz
dv/dz = yx

and the gradient of g for all x,y,z are 1.
Then I have the constraint saying: x+y+z=k

Now I use Lagranges multiplier:
GradientV = lambda*gradientG.
for x: yz = lambda
for y: xz = lambda
for z: yx = lambda

Now im lost of where to go, can anybody help me?

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Next step: yz = lambda = xz, so yz = xz. The maximum volume isn;t going to be zero, so z will not be zero and you can divide by z to get x=y. Now do the same with the other two pairs of equations. – Flounderer Mar 21 '13 at 9:54
so we got: y=x, y=z and x=z - so we basically end up with x+x+x = k? y+y+y =k? z+z+z=k? so the maximum will be when x,y,z are the same? – user1090614 Mar 21 '13 at 10:00
Exactly. The reason why you would think of doing this is because you want to eliminate the Lagrange multiplier. It is necessary to introduce it, but after that, the way to make progress is usually to get rid of it as soon as possible. – Flounderer Mar 21 '13 at 10:02
up vote 0 down vote accepted

The equations $xy = \lambda$, $yz = \lambda$ and $zx = \lambda$ can be combined to get $$ xy = yz = yx.$$ If one of the variables is not zero you get that the other two have to be equal. And if two of the variables are nonzero all of them have to be equal. Taking into account the constraint you end up with $$ (x,y,z) \in \left\{\left(0,0,k\right),\left(0,k,0\right),\left(k,0,0\right),\left(\frac{k}{3},\frac{k}{3},\frac{k}{3}\right)\right\}.$$

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