극값과 안장점

$\left(\frac{1}{4},\frac{1}{2}\right)$ is a saddle point and $(1,1)$ is the relative minimum.

$x=3$ and $y=6$

$V=\frac{64,000}{\pi }\approx 20,372$ cm³

Write $f(x,y) = x^2 +cxy +y^2$. Then

$$\begin{aligned} f_x(x,y)&= 2x+cy & f_x(0,0) = 0 \\ f_y(x,y)&= cx+2y & f_y(0,0) = 0 \\ f_{xx}(x,y) &= 2 \\ f_{xy}(x,y) &= c \\ f_{yy}(x,y) &= 2 \end{aligned}$$

As $f_x(0,0)=f_y(0,0)=0$, we have that $(0,0)$ is always a critical point for $f$. According to the Second Derivative Test, $(0,0)$ is also a saddle point for $f$ if

$$\begin{aligned} f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 <0 \iff 4-c^2 <0 \iff |c|>2 \end{aligned}$$

As a remark, the Second Derivative Test provides no information when the expression $f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 =0$, i.e. when $c=\pm 2$. But when $c=\pm 2$,

$$f(x,y) = x^2 \pm 2xy + y^2 =(x\pm y)^2$$

and $f$ has a local minimum, not a saddle point, at $(0,0)$.

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

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

The coldest point must be either on the boundary of the plate or in the interior of the plate.

• On the semi–circular part of the boundary $0\le y\le 2$ and $x^2+y^2=4$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln 5-y$. The smallest value of $\ln 5-y$ is taken when $y$ is as large as possible, i.e. when $y=2$, and is $\ln 5 -2\approx -0.391$.
• On the flat part of the boundary, $y=0$ and $-2\le x\le 2$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln\big(1+x^2\big)$. The smallest value of $\ln\big(1+x^2\big)$ is taken when $x$ is as small as possible, i.e. when $x=0$, and is $0$.
• If the coldest point is in the interior of the plate, it must be at a critical point of $T(x,y)$. Since

$$\begin{aligned} T_x(x,y)=\frac{2x}{1+x^2+y^2}\qquad T_y(x,y)=\frac{2y}{1+x^2+y^2}-1 \end{aligned}$$

a critical point must have $x=0$ and $\frac{2y}{1+x^2+y^2}-1=0$, which is the case if and only if $x=0$ and $2y-1-y^2=0$. So the only critical point is $x=0,\ y=1$, where $T=\ln 2-1\approx -0.307$.

Since $-0.391<-0.307<0$, the coldest temperture is $-0.391$ and the coldest point is $(0,2)$.

(a) The maximum and minimum can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle $x^2+y^2=4$.

• Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of $h$. Since $T_x=-8x$ and $T_y=-2y$, the only critical point is $(x,y)=(0,0)$, where $T=20$. Since $4x^2+y^2\ge 0$, we have $T(x,y)=20-4x^2-y^2\le 20$. So the maximum value of $T$ (even in $\mathbb{R}^2$) is $20$.
• At each point of $x^2+y^2=4$ we have $T(x,y)=20-4x^2-y^2=20 -4x^2-(4-x^2)=16-3x^2$ with $-2\le x\le 2$. So $T$ is a minimum when $x^2$ is a maximum. Thus the minimum value of $T$ on the disk is $16-3(\pm 2)^2=4$.

So all together, the maximum and minimum values of $T(x,y)$ in $x^2+y^2\le 4$ are $20$ (at $(0,0)$) and $4$ (at $(\pm 2,0)$), respectively.

(b) To increase its temperature as quickly as possible, the ant should move in the direction of the temperature gradient ${ \boldsymbol{\nabla} } T(1,1)=\left< -8x,-2y\right>\Big|_{(x,y)=(1,1)}=\left< -8,-2\right>$. A unit vector in that direction is $\frac{1}{\sqrt{17}}\left< -4,-1\right>$.

(c) The ant's rate f increase of temperature (per unit time) is

$$\begin{aligned} { \boldsymbol{\nabla} } T(1,1)\cdot\left< -2,-1\right> =\left< -8,-2\right>\cdot\left< -2,-1\right> =18 \end{aligned}$$

(d) We are being asked to find the $(x,y)=(x,2-x^2)$ which maximizes

$$\begin{aligned} T\big(x,2-x^2\big) =20 -4x^2-\big(2-x^2\big)^2 = 16-x^4 \end{aligned}$$

The maximum of $16-x^4$ is obviously $16$ at $x=0$. So the ant should go to $\big(0,2-0^2\big)=(0,2)$.