Find three positive numbers $x$, $y$, $z$ whose sum is $10$ such that $x^2y^2z$ is a maximum

Question

I'm self learning from the Vector Calculus book available online in PDF form (page 88). The question is: Find three positive numbers $x$, $y$, $z$ whose sum is $10$ such that $x^2y^2z$ is a maximum. Searching around for similar questions I see Lagrange multipliers mentioned as helpful, but that is subsequent material in the book so not meant to be applied here. Here's what I've done:

Rewrite the product as a function of two variables $x$ and $y$ because we know $z = 10 - x - y$:

$$f(x,y) = x^2y^2(10-x-y)$$

Compute $\frac{\partial{f}}{\partial{x}} = xy^2(20-3x-2y)$ and $\frac{\partial{f}}{\partial{y}} = x^2y(20-2x-3y)$

Now I know I need to set these to $0$ and solve simultaneously then look at the possible values of $x$ and $y$. I'm stuck here as I don't know how to solve this system:

$$xy^2(20-3x-2y) = 0$$ $$x^2y(20-2x-3y) = 0$$

I thought as they're both equal to $0$ then they're both equal to each other. Then I thought I could cancel out $x$ and $y$ but then figured I'd be losing solutions. Perhaps someone can tell me a good way to approach this?

Answer 1

As you know $x,y$ are both not zero, you can divide by them, giving $2x+3y=20=3x+2y$ Subtracting gives $x-y=0,$ so $x=y=4, z=2$

Answer 2

A solution without calculus: By the AM-GM inequality, we have $$x \cdot x \cdot y \cdot y \cdot 2z \leq \left( \frac{x+x+y+y+2z}{5} \right)^5 = 4^5 = 2^{10},$$ hence $x^2y^2z$ is at most $2^9$. Equality holds for $x=4$, $y=4$ and $z=2$.

Answer 3

Your solution is perfect. Non of $x$ and $y$ could be $0$, since then the multiplication will be $0$ as well, so with this assumption that non of $x$ and $y$ is $0$ is zero then we cancel out and $xy^2$ and $x^2y$ and we have these equations:

$2y+3x=20$

$3y+2x=20$

which implies $x=y=4$

by this we have $z=2$ and hence we have the desired multiplication to be equal to 512.