Question

1. for the following functions find all of the critical points and then classify them using the second - derivative test.

(a) (f(x,y)=x^{3}-2xy + 8y^{3}).
(b) (g(x,y)=3x^{2}+y^{3}-6xy).
(c) (h(x,y)=x^{2}y + y^{3}-x^{2}-3y^{2}).
(d) (f(x,y)=e^{(\frac{y^{3}}{3}-x - y^{2})}).
(e) (g(x,y)=8xy+\frac{16}{x}+\frac{32}{y}).

Explanation:

Step1: Find first - order partial derivatives

For a function $z = f(x,y)$, the first - order partial derivatives are $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. Critical points occur where $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$.

Step2: For $f(x,y)=x^{3}-2xy + 8y^{3}$

$\frac{\partial f}{\partial x}=3x^{2}-2y$ and $\frac{\partial f}{\partial y}=-2x + 24y^{2}$. Set $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. From $\frac{\partial f}{\partial x}=0$, we have $y=\frac{3}{2}x^{2}$. Substitute into $\frac{\partial f}{\partial y}=0$: $-2x + 24(\frac{3}{2}x^{2})^{2}=0$, $-2x+24\times\frac{9}{4}x^{4}=0$, $-2x + 54x^{4}=0$, $2x( - 1+27x^{3})=0$. So $x = 0$ or $x=\frac{1}{3}$. When $x = 0$, $y = 0$; when $x=\frac{1}{3}$, $y=\frac{1}{6}$.

Step3: Find second - order partial derivatives

$\frac{\partial^{2}f}{\partial x^{2}} = 6x$, $\frac{\partial^{2}f}{\partial y^{2}}=48y$, $\frac{\partial^{2}f}{\partial x\partial y}=-2$.

Step4: Use the second - derivative test

The discriminant $D=\frac{\partial^{2}f}{\partial x^{2}}\frac{\partial^{2}f}{\partial y^{2}}-(\frac{\partial^{2}f}{\partial x\partial y})^{2}=288xy - 4$.
For the point $(0,0)$, $D=-4<0$, so it is a saddle point.
For the point $(\frac{1}{3},\frac{1}{6})$, $\frac{\partial^{2}f}{\partial x^{2}} = 2>0$, $D=288\times\frac{1}{3}\times\frac{1}{6}-4=16 - 4=12>0$, so it is a local minimum.

We would repeat the above steps for functions (b), (c), (d) and (e) in a similar fashion.

Answer:

The process to find critical points and classify them for a function $z = f(x,y)$ involves finding first - order partial derivatives to get critical points, then second - order partial derivatives to use the second - derivative test. For $f(x,y)=x^{3}-2xy + 8y^{3}$, the critical points are $(0,0)$ (saddle point) and $(\frac{1}{3},\frac{1}{6})$ (local minimum). Similar steps are applied to other functions.