Lagrange Multiplier Question and my attempt

Question

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

$yz=xz=xy$

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

Answer 1

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

Without using Lagrange Multiplier

Assuming $x,y,z\gt0$

Using AM-GM

$$\frac{x+y+z}{3}\ge\sqrt[3]{xyz}$$

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

Answer 2

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 $x = y = z = a/3$.