극값과 안장점

(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 $(\pm 2\sqrt{3},-1)$.