Question is "Find the extrema of $xyz$ when $x+y+z=a$, $a>0$".

Starting with usual Lagrange Multiplier method, I get

$f_x = yz + \lambda = 0$

$f_y = xz + \lambda = 0$

$f_z = yz + \lambda = 0$

Now from three equations above, I multiply first by $x$ and second by $y$ and third by $z$, I get

$f_x = xyz + \lambda x = 0$

$f_y = xyz + \lambda y = 0$

$f_z = xyz + \lambda z = 0$

Clearly from these equating values of $xyz$. I get $x=y=z$. And thus I have solved the question and is consistent with my answer with textbook.

BUT, if I manipulate equations in a way as if I equate values of $\lambda$ I get


Now I take $yz=xz$. This implies either $z=0$ or $x=y$.

If I take $z=0$ and put in other two equations I get $xy=0$ which means either $x=0$ or $y=0$. Say I take $x=0$ and now putting in constraint equation I get $y=a$ so i get $(0,a,0)$. Not only this but by solving other equations like this I get $(0,a,0)$, $(a,0,0)$, $(0,0,a)$, $((a-1)/2,(a-1)/2,1)$. But this is not consistent with textbook.

Can anybody help me out from here? Thanks

  • $\begingroup$ Your working and answers look good to me. I suggest you carefully check the question in the book. It may have stated that $x,y,z,a$ are all positive (or something similar). Then the solutions $x=0$ etc would not be valid. In this case $xyz$ would have no minimum, because it can be as close as you like to zero but can never equal zero. $\endgroup$
    – David
    Nov 15, 2014 at 5:40
  • $\begingroup$ @David they have said $a > 0$ $\endgroup$ Nov 15, 2014 at 5:43
  • $\begingroup$ If $x,y,z$ are allowed to be negative then $xyz$ can be as large as you like in the negative direction and so there is no minimum. You should check the possibilities $x=0$ etc (as you have done) but in this case they do not give maximum or minimum values. Remember that solving the Lagrange equations gives possible extreme points, but some of them may in fact turn out to be not extreme points. $\endgroup$
    – David
    Nov 15, 2014 at 5:48
  • $\begingroup$ @David .then wat about $((a−1)/2, (a−1)/2, 1)$ . The correct answer is at $(a/3,a/3/a/3)$ which i have found by my former manipulation of equations . but that is fine .Out of 4 points i have obtained the first three gives $f=0$ .so they can also be ruled out because of $(a/3,a/3/a/3)$ (it gives $a^{3}/27$) . $\endgroup$ Nov 15, 2014 at 5:53
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    $\begingroup$ Doesn't matter, you are looking for maximum and minimum values, and if $f$ can be negative then $0$ isn't a minimum. $\endgroup$
    – David
    Nov 15, 2014 at 6:08

2 Answers 2


There must be some constraints on $x,y,z$

Without using Lagrange Multiplier

Assuming $x,y,z\gt0$

Using AM-GM


$$x+y+z=a\ge3\sqrt[3]{xyz}$$ $$\frac{a^3}{27}\ge xyz$$ and equality occurs when $x=y=z$

  • $\begingroup$ this is another approach .thanks $\endgroup$ Nov 15, 2014 at 6:31

You have derived the correct result with the Lagrange Multiplier method:

$xy = yz = zx = -\lambda$

Here $\lambda$ is a non-zero parameter. From this it follows that $x, y$ and $z$ can not be zero. The only solution is $x = y = z = \sqrt{-\lambda}$. And therefore $x = y = z = a/3$.

  • $\begingroup$ i see . thankss $\endgroup$ Nov 15, 2014 at 7:19

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