극값과 안장점

The specified temperature and its first order derivatives are

$$\begin{aligned} T(x,y)&=(x+y)e^{-x^2-y^2}\\ T_x(x,y)&=(1-2x^2-2xy)e^{-x^2-y^2}\\ T_y(x,y)&=(1-2xy-2y^2)e^{-x^2-y^2} \end{aligned}$$

• First, we find the critical points.

$$\begin{aligned} T_x&=0 & &\quad\iff\quad & 2x(x+y)&=1 \\ T_y&=0 & &\quad\iff\quad & 2y(x+y)&=1 \end{aligned}$$

As $x+y$ may not vanish, this forces $x=y$ and then $x=y=\pm\tfrac{1}{2}$. So the only critical points are $(\tfrac{1}{2},\tfrac{1}{2})$ and $(-\tfrac{1}{2},-\tfrac{1}{2})$.

• On the boundary $x=\cos\theta$ and $y=\sin\theta$, so $T=(\cos\theta+\sin\theta)e^{-1}$. This is a periodic function and so takes its max and min at zeroes of $\frac{\mathrm{d}T}{\mathrm{d}\theta}=\big(-\sin\theta+\cos\theta\big)e^{-1}$. That is, when $\sin\theta=\cos\theta$, which forces $\sin\theta=\cos\theta=\pm\frac{1}{\sqrt{2}}$.

All together, we have the following candidates for max and min.

$$\begin{array}{|c|c|c|c|c|}\hline \text{point} & (\tfrac{1}{2},\tfrac{1}{2}) & (-\tfrac{1}{2},-\tfrac{1}{2}) & (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) & (-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) \\ \hline \text{value of }T & \frac{1}{\sqrt{e}}\approx0.61 & -\frac{1}{\sqrt{e}} & \frac{\sqrt{2}}{e}\approx0.52 & -\frac{\sqrt{2}}{e} \\ \hline & \text{max} & \text{min} & & \\ \hline \end{array}$$

The largest and smallest values of $T$ in this table are

$$\text{min}=-\frac{1}{\sqrt{e}}\qquad \text{max}=\frac{1}{\sqrt{e}}$$