I have a problem here that I can't solve it. Can anyone help me out here, please?
Find extrema of function $f(x,y,z)= x^{2} yz$ over the unit sphere $x^{2}+y^{2}+z^{2} = 1$.
Here is my take on it: $$\begin{equation} \begin{aligned}\bigtriangledown f & = \:< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}>\\ & =\: < 2xyz, x^{2}z, x^{2} y> \end{aligned} \end{equation}$$ $$ \begin{equation} \begin{aligned}\bigtriangledown g & = \:< \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}>\\ & =\: < 2x, 2y, 2z> \end{aligned} \end{equation}$$ since there are parallel , so the cross product of them equals 0. Now, to use the conventional Lagrange Multiplier method, I get : $$ \bigtriangledown f = \lambda \ast \bigtriangledown g $$ $$\begin{equation} \begin{aligned} \begin{cases} & 2xyz= \lambda \ast 2x\\ & x^{2} z=\lambda \ast 2y \\ & x^{2} y= \lambda \ast 2z \end{cases} \end{aligned} \end{equation}$$ I can't solve x, y , and z from the system. Then I am thinking, what if I factor the common factor from both gradient vectors. $$\begin{aligned} \frac{\bigtriangledown f }{x^{2} y z} & = \;<\frac{2}{x},\frac{1}{y},\frac{1}{z}> \\ \frac{\bigtriangledown g }{2} & = \;<x, y, z> \\ \frac{\bigtriangledown f }{x^{2} y z}\times \frac{\bigtriangledown g }{2} & = <\frac{z}{y}-\frac{y}{z},\frac{x}{z}-\frac{2z}{x},\frac{2y}{x}-\frac{x}{y}> \text{Cross Product} \end{aligned}$$ $$\begin{cases} & \frac{z}{y}-\frac{y}{z}=0 \\ & \frac{x}{z}-\frac{2z}{x}=0 \\ & \frac{2y}{x}-\frac{x}{y}=0 \end{cases}$$ Simplify further: $$ \begin{cases} & \frac{z}{y}=\frac{y}{z} \\ & \frac{x}{z}=\frac{2z}{x} \\ & \frac{2y}{x}=\frac{x}{y} \end{cases}$$ Assume: \begin{aligned} \frac{y}{z} &= a\Rightarrow a = \frac{1}{a}\Rightarrow a^{2}= 1\Rightarrow a = \pm 1 \\ \frac{z}{x} &= b\Rightarrow 2b = \frac{1}{b}\Rightarrow b^{2}= \frac{1}{2}\Rightarrow b = \pm \sqrt{\frac{1}{2}}\\ \frac{y}{x} &= c\Rightarrow 2c = \frac{1}{c}\Rightarrow c^{2}= \frac{1}{2}\Rightarrow c = \pm \sqrt{\frac{1}{2}}\\ \end{aligned} Then, I have no clue what to do afterwards?
Actually, you should have four equations$$\left\{\begin{array}{l}2xyz=2\lambda x\\x^2z=2\lambda y\\x^2y=2\lambda z\\x^2+y^2+z^2=1.\end{array}\right.\label{a}\tag1$$Suppose that $\lambda=0$. Then \eqref{a} becomes$$\left\{\begin{array}{l}2xyz=0\\x^2z=0\\x^2y=0\\x^2+y^2+z^2=1.\end{array}\right.$$whose solutions are $(\pm1,0,0)$ and all triplets of the form $\left(0,\pm\sqrt{1-z^2},z\right)$.
Now, suppose that $\lambda\ne0$. It follows from the first equation of \eqref{a} that $x=0$ or that $yz=\lambda$. But, in fact, you can't have $x=0$, because then it would follow from the second and the third equations and from the fact that $\lambda\ne0$ that $y=z=0$. But then $x^2+y^2+z^2=0\ne1$.
So, $yz=\lambda$. Besides, it follows from the second and the third equations of \eqref{a} that $x^2(y+z)=2\lambda(y+z)$ and that therefore $x^2=2\lambda$ or that $y+z=0$.
If $x^2=2\lambda$, then the second equation of \eqref{a} and the fact that $\lambda\ne0$ tell us that $y=z$. But $yz=\lambda$, and therefore $y=z=\pm\sqrt\lambda$. But then the fourth equation of \eqref{a} becomes $4\lambda=1$, which means that $\lambda=\frac14$. So, we have four new solutions: $\left(\frac1{\sqrt2},\frac12,\frac12\right)$, $\left(-\frac1{\sqrt2},\frac12,\frac12\right)$, $\left(\frac1{\sqrt2},-\frac12,-\frac12\right)$, and $\left(-\frac1{\sqrt2},-\frac12,-\frac12\right)$.
And if $y+z=0$, that if, if $z=-y$, then the second equation of \eqref{a} becomes $-x^2y=2\lambda y$, and therefore $y=0$ or $x^2=-2\lambda$. But $y=0\implies z(=-y)=0$ and then it follows from the first equation of \eqref{a} and from the fact that $\lambda\ne0$ that $x=0$. But, again, that's impossible. So, $x^2=-2\lambda$ (and so $\lambda<0$). But now $2xyz=2\lambda x\implies yz=\lambda$, and so $y=\sqrt{-\lambda}$ and $z=-\sqrt{-\lambda}$ or $y=-\sqrt{-\lambda}$ and $z=\sqrt{-\lambda}$. It is easy to see that this leads to four new solutions: $\left(\frac1{\sqrt2},\frac12,-\frac12\right)$, $\left(-\frac1{\sqrt2},\frac12,-\frac12\right)$, $\left(\frac1{\sqrt2},-\frac12,\frac12\right)$, and $\left(-\frac1{\sqrt2},-\frac12,\frac12\right)$.
So, the maximum is $\frac18$, which is attained at $\left(\frac1{\sqrt2},\frac12,\frac12\right)$, at $\left(-\frac1{\sqrt2},\frac12,\frac12\right)$, at $\left(\frac1{\sqrt2},-\frac12,-\frac12\right)$, and at $\left(-\frac1{\sqrt2},-\frac12,-\frac12\right)$.