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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.

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it's $(1,2,2)$. Typing sol – Simon Markett May 24 '12 at 19:03
You missed a solution of the system of two equations. Not surprising, they are messy equations. – André Nicolas May 24 '12 at 19:05
up vote 2 down vote accepted

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 $x=1$.

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I see. But, how do you come to a conclusion that if $y$ and $z$ are positive then the equations simplify in that way? – Koba May 24 '12 at 19:32
@Dostre :The kicker is that we can write them in factored form as $(10-3y-2z)yz^2=0$ and $(10-2y-3z)y^2z=0$, so if we take $y,z$ to be non-$0$ (we don't actually need to assume positive), then we may divide the two equations by $yz^2,y^2z$ (respectively) to get the simplified equations given by Simon. Now, of course, we are looking for a solution in the first octant, so if the solution to that system of equations didn't have $y,z$ positive, then it wouldn't actually be the solution we wanted, at all. – Cameron Buie May 24 '12 at 20:40
Yeah. Now I understand it completely. Thanks. – Koba May 25 '12 at 4:58

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