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{x^{3}}{3}-x - y^{2})})
(e) (g(x,y)=8xy+\frac{16}{x}+\frac{32}{y})

Explanation:

Step1: Recall critical - point formula

For a function $z = f(x,y)$, critical points are found by setting $f_x=0$ and $f_y = 0$, where $f_x=\frac{\partial f}{\partial x}$ and $f_y=\frac{\partial f}{\partial y}$.

Step2: Differentiate $f(x,y)=x^{3}-2xy + 8y^{3}$ with respect to $x$

$f_x = 3x^{2}-2y$. Set $f_x = 0$, so $3x^{2}-2y=0\Rightarrow y=\frac{3}{2}x^{2}$.

Step3: Differentiate $f(x,y)$ with respect to $y$

$f_y=-2x + 24y^{2}$. Set $f_y = 0$, so $-2x + 24y^{2}=0$.

Step4: Substitute $y=\frac{3}{2}x^{2}$ into $-2x + 24y^{2}=0$

$-2x+24(\frac{3}{2}x^{2})^{2}=0$.
$-2x + 24\times\frac{9}{4}x^{4}=0$.
$-2x+54x^{4}=0$.
$2x( - 1 + 27x^{3})=0$.
This gives two cases:
Case 1: $x = 0$. When $x = 0$, from $y=\frac{3}{2}x^{2}$, $y = 0$.
Case 2: $27x^{3}-1=0\Rightarrow x^{3}=\frac{1}{27}\Rightarrow x=\frac{1}{3}$. When $x=\frac{1}{3}$, $y=\frac{3}{2}\times(\frac{1}{3})^{2}=\frac{1}{6}$.
So the critical points of $f(x,y)$ are $(0,0)$ and $(\frac{1}{3},\frac{1}{6})$.

Step5: Recall second - derivative test formula

The second - derivative test uses $D=f_{xx}f_{yy}-(f_{xy})^{2}$, where $f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}$, $f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}$, and $f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}$.
For $f(x,y)=x^{3}-2xy + 8y^{3}$:
$f_{xx}=6x$, $f_{xy}=-2$, $f_{yy}=48y$.
For the point $(0,0)$:
$D=(6\times0)\times(48\times0)-(-2)^{2}=- 4<0$. So $(0,0)$ is a saddle point.
For the point $(\frac{1}{3},\frac{1}{6})$:
$f_{xx}=6\times\frac{1}{3}=2$, $f_{yy}=48\times\frac{1}{6}=8$, $f_{xy}=-2$.
$D=(2)\times(8)-(-2)^{2}=16 - 4=12>0$ and $f_{xx}=2>0$. So $(\frac{1}{3},\frac{1}{6})$ is a local minimum.

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

Answer:

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). The process for other functions is similar as shown above.