Can anyone help me solve the problem below? This is question number 14.8.42 in the seventh edition of Stewart Calculus.
Here is the problem definition:
"Find the maximum and minimum volumes of a rectangular box whose surface area is $1500 cm^3$ and whose total edge length is 200 cm."
Here is my work so far(With edits based on Phyra's suggestions below):
Subject to two constraints:
Surface area: $g(x,y,z)=2xy+2xz+2yz=1500$
Edge length: $h(x,y,z)=4x+4y+4z=200$
Simplify constraints to:

Solve system of equations:
$f_x=yz=\lambda (y+z)+\mu$
$f_y=xz=\lambda (x+z)+\mu$
$f_z=xy=\lambda (x+y)+\mu$
This leads to:
$\lambda y+\lambda z+\mu=yz$
$\lambda x+\lambda z+\mu=xz$
$\lambda x+\lambda y+\mu=xy$
$\lambda x=xy-\lambda y-\mu=xz-\lambda z-\mu$, so that $xy-\lambda y=xz-\lambda z$, which simplifies to: $x(y-z)=\lambda (y-z)$, so that $\lambda=x$ when $(y-z)\ne 0$
$\lambda y =yz-\lambda z -\mu=xy-\lambda x -\mu$, so that $y(z-x)=\lambda (z-x)$, and $\lambda = y$ when $(z-x)\ne 0$
$\lambda z = yz-\lambda y -\mu=xz-\lambda x - \mu$, so that $z(y-x)=\lambda (y-x)$, and $\lambda = z$ when $(y-x)\ne 0$
The above can be summarized as:
$\lambda=x=y=z $
Substitute x=y into constraints:
$x^2 +2xz =750$ and $2x+z=50$
Substitute $z=50-2x$ into first constraint:
$3x^2-100x+750=0$ gives $x=y=\frac{5}{3}(\frac{+}{}\sqrt{10}+10)$
Also, $z=50-2x=\frac{50\frac{+}{}10\sqrt{10}}{3}$
Thus, $V=xyz\approx 2948$ or $\approx 3534$

My remaining questions are as follows:
a) What do you mean in your comments below when you refer to checking the border? This concept is not in my textbook.
b) Why choose x=y? Are you saying that choosing y=z would have given the same result?
c) Why use $z=50-2x$ instead of $x=y=z$?
d) When you mention 8 cases, to what are you referring? I do not see any cases.
e) What do you mean when you use the word symmetric in this situation?

  • $\begingroup$ You should give more details on "simplification leads to". Also, you should always examinate the border of your region. This makes it necessary to first know what your region is. $\endgroup$
    – Phira
    Nov 1, 2012 at 0:58
  • $\begingroup$ @Phira, Thank you. I elaborated to fill in detail to replace "simplification leads to", based on your request. But I do not know how to define/examine the border of my region. Do my changes above make it easier for you to help find an answer? $\endgroup$
    – CodeMed
    Nov 1, 2012 at 1:38
  • $\begingroup$ You cannot just divide by $y-z$ because it could be zero which gives several cases. $\endgroup$
    – Phira
    Nov 1, 2012 at 12:08
  • $\begingroup$ Your edit is wrong at "The above can be summarized as". You cannot just assume that $x=y=z$. $\endgroup$
    – Phira
    Nov 1, 2012 at 23:18
  • 1
    $\begingroup$ I've added details in my answer. $\endgroup$
    – Phira
    Nov 2, 2012 at 10:38

2 Answers 2


1) You really should have simplified your conditions first by dividing out 2 and 4 before differentiating it. This does not change the result, but it is needless complication.

2) $x$, $y$ and $z$ are positive numbers, so we have to take into account the behaviour on the border when one or more of them are zero.

The situation is symmetric, so say $z=0$, but then you have $x+y=50$ and $xy= 750$ which gives (by the solution formula of the quadratic equation) that $x$ and $y$ are not real numbers.

This means that subject to the two conditions there is actually no border and every extrema can be found by using Lagrange multipliers.

3) You write:

$x(y−z)/2=λ(y−z)$, so that $λ=x/2$

But what you get, for each of the three similar equations, is:

Either (a) $y=z$ or (b) $\lambda=x/2$. Either (a) $z=x$ or (b) $\lambda=y/2$. Either (a) $x=y$ or (b) $\lambda=z/2$.

Now, you would have to choose (a) or (b) for each of the three equations which gives 8 cases (aaa,aab,aba,abb,baa,bab,bba,bbb), but it is simpler than that.

If you choose at least once condition (a), then you have two equal variables. If you choose at least twice condition (b), then you have two variables equal to $2\lambda$, so you also have two equal variables.

So, whatever you do, two variables will be equal.

But the problem statement is symmetric in $x$, $y$ and $z$ (which means that if you interchange the three variables, the problem statement does not change, for example $yzx=xyz$, so the volume you are looking for will not depend on the order of the variables).

Therefore, since you know that two variables are equal and you know that it doesn't really matter for the end result which ones are equal, it suffices to investigate $x=y$:

(At this point, we still might have $x=y=z$, but we don't know that. What we do know, is that the original conditions are true, so we use them first, and it turns out that they already fix all the values.)

The original conditions give:

$2x+z=50$ and $x^2+2xz=750$.

Substituting $z=50-2x$ in the second equation gives a quadratic equation $3x^2-100x+750=0$.

This gives $x=\frac{50\pm 5\sqrt{10}}{3}$.

You find two solutions (up to symmetry):

$x=y=\frac53 (10 + \sqrt{10})$, $z=\frac{10}3(5-\sqrt{10})$ giving the volume $\frac{2500}{27}(35-\sqrt{10})\approx 2948$


$x=y=\frac 53 (10- \sqrt{10})$, $z=\frac{10}3(5+\sqrt{10})$ giving the volume $\frac{2500}{27}(35+\sqrt{10})\approx 3534$.

Since these are the only two candidates for extrema, the first one is the minimum value and the second one is the maximum value.

  • $\begingroup$ +1 for writing such a thorough and helpful reply. I have carefully reviewed your suggestions, and I have edited my original posting above to include your suggestions. However, I do not understand a number of concepts that you raised. As such, I still do not understand how the resulting answers are correct. Are you willing to answer my revised questions, which are stated at the end of my revised original posting above? $\endgroup$
    – CodeMed
    Nov 1, 2012 at 22:23
  • $\begingroup$ I am marking this as the answer. Thank you again for helping me. I am sorry for the delay. I have not been on this site in a while, but I will continue to use your advice here as a reference when working on these types of problems. $\endgroup$
    – CodeMed
    Dec 9, 2012 at 3:49

I'm also trying to figure this out and thought that x=y=z however that cannot be the case when you apply the constraints. When you use your equation for surface area where x=y=z = (50/3) you will find the area to not be equal to 1500

  • $\begingroup$ Thanks for the catch. In my own scratch paper, I just plugged $x=y=z=2\lambda$ into the surface area equation and got $\lambda=\frac{+}{}\frac{5\sqrt{10}}{2}$, which does not satisfy either the surface area equation or the edge equation. Are you able to suggest more of a solution? I just added a number of lines in my original post above to show the work I did in simplifying. Perhaps my elaborated work above can spark some ideas... $\endgroup$
    – CodeMed
    Nov 1, 2012 at 2:01

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