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


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
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Andrew Thompson

Updated on January 25, 2022


  • Andrew Thompson
    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
      Git Gud over 9 years
      The "Subject to $g(x,y,z) = 2(xy+xz+yz) - 64$" is non-sensical, 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
      user247327 about 2 years
      I 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
    Andrew Thompson over 9 years
    I made an edit according to your instructions.
  • 5xum
    5xum over 9 years
    @AndrewThompson I edited my answer.
  • colormegone
    colormegone over 9 years
    Alternatively, you can solve all three equations for $ \ \lambda \ $ to write $$ \lambda \ = \ \frac{yz}{2 \ (y+z)} \ = \ \frac{xz}{2 \ (x+z)} \ = \ \frac{xy}{2 \ (x+y)} \ \ . $$ Cross-multiplying 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.