If $x+y+z=2$, what is the max value of $xyz^2$?

Question

I have tried to find the max value of $xyz^2$ if $ x+y+z=2 $ using variation of $f(x,y,z)$, but I don't know how I can calculate the derivative of a function of three variable. If my idea is correct, then I want to ask if there is any simple way to do that.

Note: $x$, $y$, $z$ are real.

Answer 1

By AM-GM inequality twice: $xy \le \dfrac{(x+y)^2}{4} \implies xyz^2 \le \dfrac{(x+y)^2z^2}{4}= \dfrac{(2-z)^2z^2}{4} \le \dfrac{1}{4}\cdot \left(\dfrac{2^2}{4}\right)^2 = \dfrac{1}{4}$, and this is the max value you sought. It occurs at $z = 1, x = y = \dfrac{1}{2}$.

Answer 2

Note that for $x,y,z>0$ by AM-GM we have

$$\frac{x+y+\frac{z}2+\frac{z}2}4\ge \sqrt[4]{\frac{xyz^2}{4}}$$

$$\iff2=x+y+\frac{z}2+\frac{z}2\ge 4\sqrt[4]{\frac{xyz^2}{4}}$$

thus

$$xyz^2\le \frac14$$

and maximum occurs for $x=y=\frac{z}2=\frac14 \implies (x,y,z)=\left(\frac14,\frac14,\frac12\right)$.

Note also that for $xy<0 \implies f(x,y,z)<0 $.

For $x, y$ both negative let $x=y=-t$ with $t>0$ we have $z=2-(x+y)=2+2t$ and then

$$xyz^2=t^2(2+2t)^2\to +\infty$$

Therefore $f(x,y,z)$ has not maximum in $\mathbb{R^2}$ and has maximum in the restriction to $\mathbb{R^2}\setminus \{(x,y)|x<0 \land y<0\}$.

Answer 3

Note that if $x=y=10$ and $z=-18$, then $x+y+z=2$ and $f(x,y,z) = 32,400.$ There is no maximum. Using Lagrange multipliers, it's easy to find the local max at $x=y=1/2$, $z=1$.

Answer 4

The function has no maximum on that plane. Indeed, fix $y=1$, so $z=1-x$ and we have $$ f(x,1,1-x)=x(1-x)^2 $$ that has limit $\infty$ as $x\to\infty$. It has no minimum either, because the limit for $x\to-\infty$ is $-\infty$.

Answer 5

Let $$ x=t, y=-1,z=2-t$$ then $$x+y+z=1$$ and$$xyz^2= -t(2-t)^2$$ which grows without bound.

Thus There is no maximum or minimum for this function.