Lagrange Multiplier problem with three variables

Question

I struggle a bit with this problem:

Use Lagrange's methode to find the maximum value to $f(x,y,z)=xy^2z $ when $D={(x,y,z)\in R^3:x^2+y^2+z^2=36}$

So I have done this:

1. $y^2z=\lambda2x$

2. $2xyz=\lambda2y \rightarrow 0=2y(\lambda-xz)$

3. $xy^2=\lambda 2z$
4. $x^2+y^2+z^2=16$

But I'dont know what to do next, because I have not worked with 3 variables when in comes to Lagrange methode before. I thought about $y=0$ or $xz=\lambda$. But I don't know if this works. I have tried, but I always end up with some weird expressions, and never real numbers. Does anyone have some tips?

Answer 1

You have formulated the equations(1, 2, 3) correctly. Solve them to get $$\ x^2 = \lambda \\ y^2 = 2\lambda \\ z^2 = \lambda $$ Plug these in the constraint $x^2 + y^2 + z^2 = 36$. If you get multiple solutions try each solution and find which gives the maximum value. This is because Lagrangian does not always give the global maximum/minimum.

Answer 2

HINT

Indeed you have 2 cases. The first sets $y=0$ which forces the function to have zero value, and since you are looking for a maximum, this is definitely not optimal.

That forces $xz = \lambda$. Can you plug into 1st and 3rd equations and use the results to plug into the 4th?

Answer 3

Only for control:

The minimum of $$x(36-x^2-z^2)z$$

occurs at a stationary point,

$$\begin{cases}(36-3x^2-z^2)z=0,\\x(36-x^2-3z^2)=0.\end{cases}$$

The equations describe two ellipses and two straight lines, forming a total of nine intersections,

$$(x,z)=(0,0),(0,\pm6),(\pm6,0),(\pm3,\pm3).$$

The corresponding values of $f$ are $0$ and $\pm162$. The maximum occurs four times, at $(3,\pm3\sqrt2,3)$ and $(-3,\pm3\sqrt2,-3)$.