I have the following question
Using the method of the Lagrange multipliers, find the maximum and minimum values of the function $$f(x,y,z) = x^2y^2z^2$$ where $(x,y,z)$ is on the sphere $$x^2+y^2+z^2 = r^2$$ Using the result obtained to show that for arbitrary non-negative numbers $a,b$ and $c$, $$(abc)^\frac{1}{3} \leq \frac{a+b+c}{3}$$
My thoughts are to use the equation $L = f + \lambda g$, where $f$ is the same as above and $g$ is just $x^2+y^2+z^2 - r^2 = 0$
But when I compute this, for example, I get :
$L_{x} = 2xy^2z^2 + 2 \lambda x = 0$ which means that the $x$ cancel ? I think I've made a silly mistake, and was hoping someone could explain how to do this properly.
And if you can, could you guide me on the second part of the question
Thank you in advanced.
$f(x,y,z) = x^2y^2z^2$ and given constraint $g(x,y,z) = x^2+y^2+z^2-r^2 = 0$
Applying Lagrange Multiplier method,
$x^2y^2z^2 = \lambda (x^2+y^2+z^2-r^2)$
$2xy^2z^2 = 2 \lambda x$
$2x^2yz^2 = 2 \lambda y$
$2x^2y^2z = 2 \lambda z$
So if you take the first equation for example, you have $\lambda = y^2z^2, $ if $x \ne 0$
Plugging in the second and third, you get $x^2 = y^2 = z^2$ and given the constraint
$x^2 = y^2 = z^2 = \frac{r^2}{3}$
Similarly check other critical points.
EDIT: Specifically on the argument sought in the question for non-negative numbers,
Say, $a = x^2, b = y^2, c = z^2$. So they are non-negative and as per given constraint, we have
$a+b+c = r^2$
Maximum value of $\displaystyle abc = \frac{r^6}{27}$ or we say $\displaystyle abc \leq \frac{r^6}{27}$
So, $\displaystyle (abc)^{1/3} \leq \frac{r^2}{3} = \frac{a+b+c}{3}$