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: Find the first - order partial derivatives

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

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

$f_x = 3x^{2}-2y$ and $f_y=-2x + 24y^{2}$.
Set $f_x = 0$, so $3x^{2}-2y = 0\Rightarrow y=\frac{3}{2}x^{2}$.
Substitute $y=\frac{3}{2}x^{2}$ into $f_y = 0$:
$-2x+24(\frac{3}{2}x^{2})^{2}=0$.
$-2x + 24\times\frac{9}{4}x^{4}=0$.
$-2x+54x^{4}=0$.
$2x( - 1 + 27x^{3})=0$.
$2x(3x - 1)(9x^{2}+3x + 1)=0$.
The quadratic factor $9x^{2}+3x + 1$ has no real roots since $\Delta=3^{2}-4\times9\times1=9 - 36=-27<0$.
If $2x = 0$, then $x = 0$, and $y = 0$.
If $3x - 1=0$, then $x=\frac{1}{3}$, and $y=\frac{3}{2}\times(\frac{1}{3})^{2}=\frac{1}{6}$.
The critical points are $(0,0)$ and $(\frac{1}{3},\frac{1}{6})$.
The second - order partial derivatives are $f_{xx}=6x$, $f_{xy}=-2$, $f_{yy}=48y$.
The discriminant $D=f_{xx}f_{yy}-(f_{xy})^{2}=6x\times48y-(-2)^{2}=288xy - 4$.
For $(0,0)$: $D=-4<0$, so $(0,0)$ is a saddle point.
For $(\frac{1}{3},\frac{1}{6})$: $f_{xx}=6\times\frac{1}{3}=2$, $D=288\times\frac{1}{3}\times\frac{1}{6}-4=16 - 4 = 12>0$, so $(\frac{1}{3},\frac{1}{6})$ is a local minimum.

(b) For $g(x,y)=3x^{2}+y^{3}-6xy$

$g_x = 6x-6y$ and $g_y=3y^{2}-6x$.
Set $g_x = 0$, so $x = y$.
Substitute $x = y$ into $g_y = 0$:
$3y^{2}-6y=0$.
$3y(y - 2)=0$.
If $y = 0$, then $x = 0$. If $y = 2$, then $x = 2$.
The critical points are $(0,0)$ and $(2,2)$.
The second - order partial derivatives are $g_{xx}=6$, $g_{xy}=-6$, $g_{yy}=6y$.
The discriminant $D = g_{xx}g_{yy}-(g_{xy})^{2}=6\times6y-(-6)^{2}=36y - 36$.
For $(0,0)$: $D=-36<0$, so $(0,0)$ is a saddle point.
For $(2,2)$: $D=36\times2-36 = 36>0$ and $g_{xx}=6>0$, so $(2,2)$ is a local minimum.

(c) For $h(x,y)=x^{2}y + y^{3}-x^{2}-3y^{2}$

$h_x=2xy-2x=2x(y - 1)$ and $h_y=x^{2}+3y^{2}-6y$.
Set $h_x = 0$, then $x = 0$ or $y = 1$.
If $x = 0$, then $h_y=3y^{2}-6y=3y(y - 2)=0$, so $y = 0$ or $y = 2$.
If $y = 1$, then $h_y=x^{2}+3 - 6=x^{2}-3=0$, so $x=\pm\sqrt{3}$.
The critical points are $(0,0)$, $(0,2)$, $(\sqrt{3},1)$ and $(-\sqrt{3},1)$.
The second - order partial derivatives are $h_{xx}=2y - 2$, $h_{xy}=2x$, $h_{yy}=6y - 6$.
The discriminant $D=h_{xx}h_{yy}-(h_{xy})^{2}=(2y - 2)(6y - 6)-(2x)^{2}=12(y - 1)^{2}-4x^{2}$.
For $(0,0)$: $D = 12>0$ and $h_{xx}=-2<0$, so $(0,0)$ is a local maximum.
For $(0,2)$: $D = 12>0$ and $h_{xx}=2>0$, so $(0,2)$ is a local minimum.
For $(\sqrt{3},1)$: $D=-12<0$, so $(\sqrt{3},1)$ is a saddle point.
For $(-\sqrt{3},1)$: $D=-12<0$, so $(-\sqrt{3},1)$ is a saddle point.

(d) For $F(x,y)=e^{(\frac{x^{3}}{3}-x - y^{2})}$

$F_x=(x^{2}-1)e^{(\frac{x^{3}}{3}-x - y^{2})}$ and $F_y=-2ye^{(\frac{x^{3}}{3}-x - y^{2})}$.
Set $F_x = 0$, then $x^{2}-1 = 0\Rightarrow x=\pm1$.
Set $F_y = 0$, then $y = 0$.
The critical points are $(1,0)$ and $(-1,0)$.
The second - order partial derivatives:
$F_{xx}=(2x)e^{(\frac{x^{3}}{3}-x - y^{2})}+(x^{2}-1)^{2}e^{(\frac{x^{3}}{3}-x - y^{2})}$, $F_{xy}=2xye^{(\frac{x^{3}}{3}-x - y^{2})}$, $F_{yy}=(4y^{2}-2)e^{(\frac{x^{3}}{3}-x - y^{2})}$.
For $(1,0)$:
$F_{xx}=0+(0)e^{(\frac{1}{3}-1-0)} = 0$, $F_{xy}=0$, $F_{yy}=-2e^{-\frac{2}{3}}$.
$D=0 - 0=0$, the second - derivative test is inconclusive.
For $(-1,0)$:
$F_{xx}=-2e^{\frac{1}{3}}+(4)e^{\frac{1}{3}}=2e^{\frac{1}{3}}$, $F_{xy}=0$, $F_{yy}=-2e^{\frac{1}{3}}$.
$D=-4e^{\frac{2}{3}}<0$, so $(-1,0)$ is a saddle point.

(e) For $G(x,y)=8xy+\frac{16}{x}+\frac{32}{y}$

$G_x=8y-\frac{16}{x^{2}}$ and $G_y=8x-\frac{32}{y^{2}}$.
Set $G_x = 0$, so $y=\frac{2}{x^{2}}$.
Substitute $y=\frac{2}{x^{2}}$ into $G_y = 0$:
$8x-\frac{32}{(\frac{2}{x^{2}})^{2}}=0$.
$8x-\frac{32}{\frac{4}{x^{4}}}=0$.
$8x - 8x^{4}=0$.
$8x(1 - x^{3})=0$.
$x = 0$ is not in the domain. If $x = 1$, then $y = 2$.
The critical point is $(1,2)$.
The second - order partial derivatives are $G_{xx}=\frac{32}{x^{3}}$, $G_{xy}=8$, $G_{yy}=\frac{64}{y^{3}}$.
At $(1,2)$: $G_{xx}=32$, $G_{xy}=8$, $G_{yy}=8$.
$D=32\times8-64=256 - 64 = 192>0$ and $G_{xx}=32>0$, so $(1,2)$ is a local minimum.

Answer:

(a) Critical points: $(0,0)$ (saddle point), $(\frac{1}{3},\frac{1}{6})$ (local minimum)

(b) Critical points: $(0,0)$ (saddle point), $(2,2)$ (local minimum)

(c) Critical points: $(0,0)$ (local maximum), $(0,2)$ (local minimum), $(\sqrt{3},1)$ (saddle point), $(-\sqrt{3},1)$ (saddle point)

(d) Critical points: $(1,0)$ (test inconclusive), $(-1,0)$ (saddle point)

(e) Critical point: $(1,2)$ (local minimum)