找到$f$的絕對最大值和最小值

$$\ nabla f =（\ cos x， - \ sin y）$$

$$\ nabla f = 0 \ Rightarrow \ cos x = 0 \ text {和} - \ sin y = 0 \ Rightarrow x = \ frac {\ pi} {2}，\ frac {3 \ pi} {2} \ text {和} y = 0，\ pi，2 \ pi$$

$$\ frac {\ partial ^ 2 {f}} {\ partial {x ^ 2}} = - \ sin x \ \，\ \ \ frac {\ partial ^ 2 {f}} {\ partial {y ^ 2 }} = - \ cos y \ \，\ \ \ frac {\ partial ^ 2 {f}} {\ partial {x} \ partial {y}} = 0$$

• $$\left (\frac{\pi}{2}, 0\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, 0\right )=-1<0$$
• $$\left (\frac{\pi}{2}, \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, \pi\right )=-1<0$$
• $$\left (\frac{\pi}{2}, 2 \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, 2 \pi\right )=-1<0$$
• $$\left (\frac{3\pi}{2}, 0\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=-1<0$$
• $$\left (\frac{3\pi}{2}, \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{3\pi}{2}, \pi\right )=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=1>0 \Rightarrow \left (\frac{3\pi}{2}, \pi\right ) \text{ is a local minima }$$
• $$\left (\frac{3\pi}{2}, 2\pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{3\pi}{2}, 2\pi\right )=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=1>0 \Rightarrow \left (\frac{3\pi}{2}, 0\right ) \text{ is a local minima }$$



EDIT1:

1. 在$U$中查找$f$的所有關鍵點。
2. 當我們將其視為僅在$\ partial U$中定義的函數時，找到$f$的關鍵點。
3. 計算所有關鍵點的$f$值。
4. 比較所有這些值並選擇最大值和最小值。

$U$是開放的矩形。我們怎麽做第二步呢？



EDIT2:

$$\ nabla f =（\ cos x， - \ sin y）$$

$U =（0,2 \ pi）\ times（0,2 \ pi）$

$$\ nabla f = 0 \ Rightarrow \ cos x = 0 \ text {和} - \ sin y = 0 \ Rightarrow x = \ frac {\ pi} {2}，\ frac {3 \ pi} {2} \ text {和} y = \ pi$$

$U$中的關鍵點是$$\ left（\ frac {\ pi} {2}，\ pi \ right），\ left（\ frac {3 \ pi} {2}，\ pi \ right）$$

1. 在$U =（ - 1,1）\ times（-1,1）$中查找$f$的所有關鍵點。
2. 當我們將其視為僅在$\ partial U$中定義的函數時，找到$f$的關鍵點。
3. 計算所有關鍵點的$f$值。
4. 比較所有這些值並選擇最大值和最小值。

First step:

$$\ nabla f =（y，x）$$

$$\ nabla f = 0 \ Rightarrow y = 0 \ text {和} x = 0$$

$U$中唯一的關鍵點是$$（0,0）$$

Second step:

$\partial U=A\cup B\cup C\cup D$ with \begin{eqnarray*} A &=&\left\{ (x,-1):-1\leq x\leq 1 \right\} \\ B &=&\left\{ (1 ,y):-1\leq y\leq 1 \right\} \\ C &=&\left\{ (x,1 ):-1\leq x\leq 1 \right\} \\ D &=&\left\{ (-1,y):-1\leq y\leq 1 \right\} . \end{eqnarray*}

$$A：f（x，-1）= - x = g（x）\\ g'（x）= 0 \ Rightarrow -1 = 0$$ $$B：f（1，y）= y = g（y）\\ g'（y）= 0 \ Rightarrow 1 = 0$$ $$C：f（x，1）= x = g（x）\\ g'（x）= 0 \ Rightarrow 1 = 0$$ $$D：f（-1，y）= - y = g（y）\\ g'（y）= 0 \ Rightarrow -1 = 0$$ 這意味著邊界$\ partial {U}$上沒有關鍵點。

Third step:

Forth step: