극값과 안장점

(a) We have

$$\begin{aligned} f(x,y)&=xy(x+2y-6)\quad & f_x(x,y)&=2xy+2y^2-6y\quad & f_{xx}(x,y)&=2y \\ & & f_y(x,y)&=x^2+4xy-6x & f_{yy}(x,y)&=4x\\ & & & &f_{xy}(x,y)&=2x+4y-6 \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff 2y(x+y-3)=0\text{ and }x(x+4y-6)=0 \\ &\iff \{y=0\text{ or }x+y=3\}\text{ and }\{x=0\text{ or }x+4y=6\}\\ &\iff \{x=y=0\}\text{ or }\{y=0,\ x+4y=6\}\\ &\text{ or }\{x+y=3,\ x=0\} \text{ or }\{x+y=3,\ x+4y=6\}\\ &\iff (x,y)=(0,0)\text{ or }(6,0)\text{ or }(0,3)\text{ or }(2,1) \end{aligned}$$

Here is a table giving the classification of each of the four critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & 0\times 0-(-6)^2<0 & & \text{saddle point} \\ \hline (6,0) & 0\times 24-6^2<0 & & \text{saddle point} \\ \hline (0,3) & 6\times 0-6^2<0 & & \text{saddle point} \\ \hline (2,1) & 2\times 8-2^2>0 & \text{2} & \text{local min} \\ \hline \end{array}$$

(b) Observe that $xy=4$ and $x+2y=6$ intersect when $x=6-2y$ and

$$\begin{aligned} (6-2y)y=4 &\iff 2y^2-6y+4=0 \iff 2(y-1)(y-2)=0 \\ &\iff (x,y) = (4,1)\text{ or }(2,2) \end{aligned}$$

The shaded region in the sketch below is $D$.

None of the critical points are in $D$. So the max and min must occur at either $(2,2)$ or $(4,1)$ or on $xy=4$, $2<x<4$ (in which case $F(x)=f\big(x,\frac{4}{x}\big)=4\big(x+\frac{8}{x}-6)$ obeys $F'(x)=4-\frac{32}{x^2}=0\iff x=\pm2\sqrt{2}$) or on $x+2y=6$, $2<x<4$ (in which case $f(x,y)$ is identically zero). So the min and max must occur at one of

$$\begin{array}{|c|c|}\hline (x,y) & f(x,y) \\ \hline (2,2) & 2\times 2(2+2\times2-6)=0 \\ \hline (4,1) & 4\times 1(4+2\times 1-6)=0 \\ \hline (2\sqrt{2},\sqrt{2}) & 4(2\sqrt{2}+2\sqrt{2}-6)<0 \\ \hline \end{array}$$

The maximum value is $0$ and the minimum value is $4(4\sqrt{2}-6) \approx -1.37$.

We have

$$\begin{aligned} f(x,y)&=x^4+y^4-4xy\qquad & f_x(x,y)&=4x^3-4y\qquad & f_{xx}(x,y)&=12x^2\\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2 \\ & & & &f_{xy}(x,y)&=-4 \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \iff x=x^9\text{ and }y=x^3 \\ &\iff x(x^8-1)=0,\ y=x^3\\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{aligned}$$

Here is a table giving the classification of each of the three critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & 0\times 0-(-4)^2<0 & & \text{saddle point} \\ \hline (1,1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline (-1,-1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline \end{array}$$

We have

$$\begin{aligned} f(x,y)&=x^3+xy^2-x\qquad & f_x(x,y)&=3x^2+y^2-1\qquad & f_{xx}(x,y)&=6x \\ & & f_y(x,y)&=2xy & f_{yy}(x,y)&=2x \\ & & & & f_{xy}(x,y)&=2y \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff xy=0\text{ and }3x^2+y^2=1 \\ &\iff \{x=0\text{ or }y=0\}\text{ and }3x^2+y^2=1 \\ &\iff (x,y)=(0,1)\text{ or }(0,-1)\text{ or }\left(\frac{1}{\sqrt{3}},0\right) \text{ or }\left(-\frac{1}{\sqrt{3}},0\right) \end{aligned}$$

Here is a table giving the classification of each of the four critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,1) & 0\times 0-2^2<0 & & \text{saddle point} \\ \hline (0,-1) & 0\times 0-(-2)^2<0 & & \text{saddle point} \\ \hline \big(\frac{1}{\sqrt{3}},0\big) & 2\sqrt{3}\times\frac{2}{\sqrt{3}}-0^2>0 & 2\sqrt{3} & \text{local min} \\ \hline \big(-\frac{1}{\sqrt{3}},0\big) & -2\sqrt{3}\times \big(-\frac{2}{\sqrt{3}}\big)-0^2>0 & -2\sqrt{3} & \text{local max} \\ \hline \end{array}$$

(a) We have

$$\begin{aligned} g(x,y)&=x^2-10y-y^2 \qquad& g_x(x,y)&=2x \qquad& g_{xx}(x,y)&=2\\ & & g_y(x,y)&=-10-2y\qquad & g_{yy}(x,y)&=-2 \\ & & & &g_{xy}(x,y)&=0 \end{aligned}$$

At a critical point

$$\begin{aligned} g_x(x,y)=g_y(x,y)=0 \iff 2x=0\text{ and }-10-2y=0 \iff (x,y)=(0,-5) \end{aligned}$$

Since $g_{xx}(0,-5)g_{yy}(0,-5)-g_{xy}(0,-5)^2=2\times(-2)-0^2<0$, the critical point is a saddle point.

(b) The extrema must be either on the boundary of the region or in the interior of the region.

• On the semi-elliptical part of the boundary $-2\le y\le 0$ and $x^2+4y^2=16$ so that $g=x^2-10y-y^2=16-10y-5y^2=21-5(y+1)^2$. This has a minimum value of 16 (at $y=0,-2$) and a maximum value of 21 (at $y=-1$). You could also come to this conclusion by checking the critical point of $16-10y-5y^2$ (i.e. solving $\frac{\mathrm{d}}{\mathrm{d}y}(16-10y-5y^2)=0$) and checking the end points of the allowed interval (namely $y=0$ and $y=-2$).
• On the flat part of the boundary $y=0$ and $-4\le x\le 4$ so that $g=x^2$. The smallest value is taken when $x=0$ and is $0$ and the largest value is taken when $x=\pm 4$ and is $16$.
• If an extremum is in the interior of the plate, it must be at a critical point of $g(x,y)$. The only critical point is not in the prescribed region.

Here is a table giving all candidates for extrema:

$$\begin{array}{|c|c|}\hline (x,y) & g(x,y) \\ \hline (0,-2) & 16 \\ \hline (\pm 4,0) & 16 \\ \hline (\pm \sqrt{12},-1) & 21 \\ \hline (0,0) & 0 \\ \hline \end{array}$$

From the table the smallest value of $g$ is $0$ at $(0,0)$ and the largest value is $21$ at $(\pm 2\sqrt{3},-1)$.

We have

$$\begin{aligned} f(x,y)&=x^3-3xy^2-3x^2-3y^2\qquad & f_x(x,y)&=3x^2-3y^2-6x\qquad & f_{xx}(x,y)&=6x-6 \\ & & f_y(x,y)&=-6xy-6y & f_{yy}(x,y)&=-6x-6 \\ & & & &f_{xy}(x,y)&=-6y \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff 3(x^2-y^2-2x)=0\text{ and }-6y(x+1)=0 \\ &\iff \{x=-1\text{ or }y=0\}\text{ and }x^2-y^2-2x=0 \\ &\iff (x,y)=(-1,\sqrt{3})\text{ or }(-1,-\sqrt{3})\text{ or }(0,0) \text{ or }(2,0) \end{aligned}$$

Here is a table giving the classification of each of the four critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & (-6)\times(-6)-0^2>0 & -6 & \text{local max} \\ \hline (2,0) & 6\times(-18)-0^2<0 & & \text{saddle point} \\ \hline (-1,\sqrt{3}) & (-12)\times0-(-6\sqrt{3})^2<0 & & \text{saddle point} \\ \hline (-1,-\sqrt{3}) & (-12)\times 0-(6\sqrt{3})^2<0 & & \text{saddle point} \\ \hline \end{array}$$

The maximum must be either on the boundary of $D$ or in the interior of $D$.

• On the circular part of the boundary $r=2$, $0\le\theta\le\frac{\pi}{2}$ (in polar coordinates) so that $f=r^2\cos\theta\sin\theta e^{-r^2/2}=2\sin(2\theta)e^{-2}$. This has a maximum value of $2e^{-2}$ at $\theta=\frac{\pi}{4}$ or $x=y=\sqrt{2}$.
• On the two flat parts of the boundary $x=0$ or $y=0$ so that $f=0$.
• If the maximum is in the interior of $D$, it must be at a critical point of $f(x,y)$. Since

$$\begin{aligned} f_x(x,y)=e^{-(x^2 + y^2) / 2}\big[y-x^2y\big]\qquad f_y(x,y)=e^{-(x^2 + y^2) / 2}\big[x-xy^2\big] \end{aligned}$$

$(x,y)$ is a critical point if and only if

$$\begin{aligned} &y(1-x^2)=0\text{ and }x(1-y^2)=0 \\ &\iff \{y=0\text{ or }x=1\text{ or }x=-1\}\text{ and } \{x=0\text{ or }y=1\text{ or }y=-1\} \end{aligned}$$

There are two critical points with $x,y\ge 0$, namely $(0,0)$ and $(1,1)$. The first of these is on the boundary of $D$ and the second is in the interior of $D$.

Here is a table giving all candidates for the maximum:

$$\begin{array}{|c|c|}\hline (x,y) & g(x,y) \\ \hline (\sqrt{2},\sqrt{2}) & 2e^{-2}\approx0.271 \\ \hline (x,0) & 0 \\ \hline (0,y) & 0 \\ \hline (1,1) & e^{-1}\approx0.368 \\ \hline \end{array}$$

Since $e>2$, we have that $2e^{-2}=e^{-1}\frac{2}{e}<e^{-1}$ and the largest value is $e^{-1}$.