극값과 안장점

The maximum and minimum must either occur at a critical point or on the boundary of $R$.

• The critical points are the solutions of

$$\begin{aligned} 0&=f_x(x,y) = 2-2x \\ 0&=f_y(x,y) = -8y \\ \end{aligned}$$

So the only critical point is $(1,0)$.

• On the side $x=-1$, $-1\le y\le 1$ of the boundary of $R$

$$f(-1,y) = 2-4y^2$$

This function decreases as $|y|$ increases. So its maximum value on $-1\le y\le 1$ is achieved at $y=0$ and its minimum value is achieved at $y=\pm 1$.

• On the side $x=3$, $-1\le y\le 1$ of the boundary of $R$

$$f(3,y) = 2-4y^2$$

This function decreases as $|y|$ increases. So its maximum value on $-1\le y\le 1$ is achieved at $y=0$ and its minimum value is achieved at $y=\pm 1$.

• On both sides $y=\pm 1$, $-1\le x\le 3$ of the boundary of $R$

$$f(x,\pm 1) = 1+2x-x^2 = 2 -(x-1)^2$$

This function decreases as $|x-1|$ increases. So its maximum value on $-1\le x\le 3$ is achieved at $x=1$ and its minimum value is achieved at $x= 3$ and $x=-1$ (both of whom are a distance $2$ from $x=1$).

So we have the following candidates for the locations of the min and max

$$\begin{array}{|c|c|c|c|c|c|c|}\hline \text{point} & (1,0) & (-1,0) & (1,\pm 1) & (-1,\pm 1) & (3,0) & (3,\pm 1) \\ \hline \text{value of }f & 6 & 2 & 2 & -2 & 2 & -2 \\ \hline & \text{max} & & & \text{min} & & \text{min} \\ \hline \end{array}$$

So the minimum is $-2$ and the maximum is $6$.