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

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.

## closed as off-topic by Namaste, José Carlos Santos, mlc, Xander Henderson, SaadMar 13 '18 at 0:31

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• are the $x,y,z$ assumed to be positive? – Dr. Sonnhard Graubner Feb 18 '18 at 10:44
• Is it $x\cdot y \cdot z^2$ or $(xyz)^2$? – Landuros Feb 18 '18 at 10:44
• Without more restrictions, I think $f(x,y,z)$ couldn't have a maximum. – Dog_69 Feb 18 '18 at 10:51
• -1 People pointed out as soon as you posted the question that you need to require $x,y,z$ non-negative. Otherwise you can take $x=-N,y=-N,z=2N+2$ and get $xyz^2=4N^2(N+1)^2$ arbitrarily large. Please edit the question. – almagest Feb 18 '18 at 19:50

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 $z = 1, x = y = \dfrac{1}{2}$.

• you can not use $AM-GM$ the $x,y,z$ are assumed to be reals – Dr. Sonnhard Graubner Feb 18 '18 at 10:46
• What is AM-GM ? – user517526 Feb 18 '18 at 10:47
• i think it is a local maximum – Dr. Sonnhard Graubner Feb 18 '18 at 10:55
• You seem to believe that $z^2(2-z^2)^2\le 1$, which is obviously false. – egreg Feb 18 '18 at 11:24
• $z^2(2-z)^2\leqslant 1$ is unjustified, if $z \in \mathbb R$. For e.g. consider $z=-10$. – Macavity Feb 18 '18 at 11:28

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 $-\infty$.

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 $\mathbb{R^2}\setminus \{(x,y)|x<0 \land y<0\}$.

• then it must be $$x>0,y>0,z>0$$! – Dr. Sonnhard Graubner Feb 18 '18 at 11:01
• Well, in fact it must be $\;xy\ge0\;$ ... – DonAntonio Feb 18 '18 at 11:05

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 $x=y=1/2$, $z=1$.

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.