Maximizing the volume of a box using Lagrange multipliers
First, write down the equations you get when differenting $L$. For example, the derivative with respect to $x$ is $$\frac{dL}{dx} = yz + 2\lambda(y+z).$$
Then, you set the derivatives to $0$. This gives you a system of $4$ equations with $4$ variables to solve. Write down that system as an edit to your question so we can see if you made a mistake up to this point.
Edit: As far as I can see, you made no error in calculating the derivatives. Now, I advise you to look at the equations
$$yz=2\lambda(y+z)\\xz=2\lambda(x+z)\\xy=2\lambda(x+y)$$
Try to multiply the first equation by $x$, the second by $y$ and the third by $z$. Play around with what you get, see where that leads you.
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Andrew Thompson
Updated on January 25, 2022Comments

Andrew Thompson almost 2 years
We are given a box of surface area $64$. As such, I wish to maximize $f(x,y,z) = xyz$ subject to $g(x,y,z) = 2(xy+xz+yz)  64$. If I have understood in correctly, I am to find the critical points of the function $L(x,y,z,\lambda) = xyz + \lambda(2xy+2xz+2yz  64)$. However, I always end up with a contradiction. Can anyone give me a thorough explanation of how one should go about attacking this problem?
We have
$L_1 = yz + \lambda(2y + 2z) = 0\\ L_2 = xz + \lambda(2x + 2z) = 0\\ L_3 = xy + \lambda(2x + 2y) = 0\\ L_4 = 2xy + 2xz + 2yz  64 = 0$
I have attempted several methods, all leave me with contradictions, and are incredibly long and tedious. The tasks are generally made so that if you are on the right track, you will get somewhat "clean" results.

Git Gud over 9 yearsThe "Subject to $g(x,y,z) = 2(xy+xz+yz)  64$" is nonsensical, it should be something like "Subject to $2(xy+xz+yz)=64$", in which case you set up the problem correctly. Lagrange multipliers do not seem to work here, there are no absolute extrema.

user247327 about 2 yearsI find that, since a specific value of $\lambda$ is not necessary for a solution, it is often best to start by eliminating $\lambda$, perhaps by DIVIDING. $yz+ \lambda(2y+ 2z)= 0$ can be written $\lambda(2y+2z)= yz$ and $xz+ \lambda(2x+ 2z)= 0$ can be written $\lambda(2x+ 2z)= xz$. Dividing, $\frac{y+ z}{x+ z}= \frac{y}{x}$ or $xy+ xz= xy+ yz$ so $xz= yz$ and, for z not 0, $x= y$. Do the same with the other equations.


Andrew Thompson over 9 yearsI made an edit according to your instructions.

5xum over 9 years@AndrewThompson I edited my answer.

colormegone over 9 yearsAlternatively, you can solve all three equations for $ \ \lambda \ $ to write $$ \lambda \ = \ \frac{yz}{2 \ (y+z)} \ = \ \frac{xz}{2 \ (x+z)} \ = \ \frac{xy}{2 \ (x+y)} \ \ . $$ Crossmultiplying the first pair (which is "safe" since none of the variable values will equal zero), we obtain $ \ 2 \ (xyz \ + \ yz^2) \ = \ 2 \ (xyz \ + \ xz^2) \ \Rightarrow \ y \ = \ x \ $ . Carrying this out for all three pairs will lead to $ \ x \ = \ y \ = \ z \ $ : perhaps unsurprisingly, this tells us that the maximal volume box is a cube.