Q:

Finding extremas of a three variable function

Find all points on he portion of the plane $x+y+z=5$ in the first octant at which $f(x,y,z)=xy^2z^2$ has a maximum value.

Attempt; Since $x+y+z=5$; $x=5-y-z$. I plug this into the $f(x,y,z)$:
$$f(5-y-z,y,z)=u(y,z)=y^2 z^2 (5-y-z)=\text{5 }y^2 z^2-y^3 z^2-y^2 z^3$$
Now I find critical points:
$$u_y=10 yz^2-3y^2z^2-2yz^3=0$$
$$u_z=10 y^2 z-2 y^3 z-3 y^2 z^2=0$$
The solution for this system of equations is $y=z=0$.Therefore, $x=5$. So thats the only critical point $(0,0,5)$ I get and $f(0,0,5)=0$. The answer should be a max at $(1,1,2)$ according to the answer key. How do I get it? Any hints please. 

A:

The maximum is at $(1,2,2)$  (I assume there is a typo, either in your question or in the answer key.)
You did everything correctly up to the point where you determine the critical points. We have
$$10yz^2-3y^2z^2-2yz^3=0$$
and
$$10y^2z-2y^3z-3y^2z^2=0$$
The solution is not only $z=y=0$, but it is a sufficient condition if either $y$ or $z$ is $0$. You also get a solution where $z,y\neq 0$.
First look at the original function. In the first octant the values are non-negative. If $y=0$ or $z=0$ the value is also $0$. So let us assume both are positive. Then the equations simplify to
$$10-3y-2z=0$$
and 
$$10-2y-3z=0$$
We see easily that the only solution is $y=z=2$ which yields $x=1$.

