Max min with $xy^2+yz^2$

Question

Find the maximum and minimum values of $xy^2+yz^2$ over the ball $x^2+y^2+z^2\le 1$.

My work: $$ x^2+y^2+z^2=1 \Rightarrow z^2=1-x^2-y^2\\ f(x,y,z)=xy^2+yz^2\Rightarrow f(x,y)=xy^2+y(1-x^2-y^2)\Rightarrow f(x,y)=xy^2+y-x^2y-y^3\\ f_{1}=y^2-2xy=0\Rightarrow y=0,2x\\ f_{2}=2xy+1-x^2-3y^2=0\Rightarrow x=\pm \frac{1}{3}\\ y=0,2x\ and\ x=\pm\frac{1}{3} \Rightarrow y=\pm \frac{2}{3} $$

How do i proceed to get $x$ and $z$, and max and min?

According to the facit the answer is max $\frac {4}{9}$ and min $-\frac {4}{9}$.

EDIT: Put the possible values of $x$ and $y$ into $x^2+y^2+z^2=1$. $$x=\frac{1}{3} \Rightarrow y=\frac{2}{3}\Rightarrow z=\pm \frac{2}{3}\\ x=-\frac{1}{3} \Rightarrow y=-\frac{2}{3}\Rightarrow z=\pm \frac{2}{3}\\$$ Put the values into $f(x,y,z)=xy^2+yz^2$, and i get max $\frac {4}{9}$ and min $-\frac {4}{9}$.

Answer 1

The first step is to find the critical points of the function $f(x,y,z) = xy^2+yz^2$ inside the unit circumference $\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2<1\}$.

So, you must compute the partial derivatives of $f$:

\begin{align} \frac{\partial f}{\partial x} =& y^2\\ \frac{\partial f}{\partial y} =& 2xy + z^2\\ \frac{\partial f}{\partial z} =& 2yz. \end{align} The critical points of $f$ are those who vanished the partial derivatives simultaneously. Then, necessarily $y=0$ because $\frac{\partial f}{\partial x} = 0$. With this we also have that $\frac{\partial f}{\partial z} =0$. Finally, we need to solve $$\frac{\partial f}{\partial x} = z^2,$$ (recall $y=0$). This compels to $z=0$. Thus, the critical points of $f$ are $$\{(x,y,z)\in\mathbb{R}^3: y=z=0, -1\leq x \leq 1\}.$$ Notice that $f(x,0,0)=0$, for all $x\in[-1,1]$.

Now it comes the hardest part, to analyze the values on the boundary. Usually this can be done using Lagrange multipliers.

Answer 2

We can use AM-GM here: $$xy^2+yz^2=y(xy+z^2)\leq|y|(|xy|+z^2)=|y|\left(\frac{1}{2}\cdot|2xy|+z^2\right)\leq$$ $$\leq|y|\left(\frac{1}{2}\cdot\frac{4x^2+y^2}{2}+z^2\right)=|y|\left(x^2+z^2+\frac{1}{4}y^2\right)\leq$$ $$\leq|y|\left(1-y^2+\frac{1}{4}y^2\right)=\sqrt{y^2\left(1-\frac{3}{4}y^2\right)^2}=\frac{3}{2}\sqrt{y^2\left(\frac{2}{3}-\frac{1}{2}y^2\right)^2}\leq$$ $$\leq\frac{3}{2}\sqrt{\left(\frac{y^2+2\left(\frac{2}{3}-\frac{1}{2}y^2\right)}{3}\right)^3}=\frac{4}{9}.$$ The equality occurs for $y^2=\frac{2}{3}-\frac{1}{2}y^2,$ $2|x|=|y|$ and $x^2+y^2+z^2=1,$ which says that we got a maximal value.

The minimal value we can get by the same way.