극값과 안장점

(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 $\big(0,2-0^2\big)=(0,2)$.

(a) For $x,y>0$,

$$\begin{aligned} f_x&=2-\frac{1}{x^2y}=0\iff y=\frac{1}{2x^2}\\ f_y&=4-\frac{1}{xy^2}=0 \end{aligned}$$

Substituting $y=\frac{1}{2x^2}$, from the first equation, into the second gives $4-4x^3=0$ which forces $x=1$, $y=\tfrac{1}{2}$. At $x=1$, $y=\tfrac{1}{2}$,

$$f\big(1,\tfrac{1}{2}\big)=2+2+2=6$$

(b) The second derivatives are

$$f_{xx}(x,y)=\frac{2}{x^3y}\qquad f_{xy}(x,y)=\frac{1}{x^2y^2}\qquad f_{yy}(x,y)=\frac{2}{xy^3}$$

In particular

$$f_{xx}\big(1,\tfrac{1}{2}\big)=4\qquad f_{xy}\big(1,\tfrac{1}{2}\big)=4\qquad f_{yy}\big(1,\tfrac{1}{2}\big)=16$$

Since $f_{xx}\big(1,\tfrac{1}{2}\big)f_{yy}\big(1,\tfrac{1}{2}\big)- f_{xy}\big(1,\tfrac{1}{2}\big)^2=4\times 16-4^2=48>0$ and $f_{xx}\big(1,\tfrac{1}{2}\big)=4>0$, the point $\big(1,\tfrac{1}{2}\big)$ is a local minimum.

(c) As $x$ or $y$ tends to infinity (with the other at least zero), $2x+4y$ tends to $+\infty$. As $(x,y)$ tends to any point on the first quadrant part of the $x$- and $y$–axes, $\frac{1}{xy}$ tends to $+\infty$. Hence as $x$ or $y$ tends to the boundary of the first quadrant (counting infinity as part of the boundary), $f(x,y)$ tends to $+\infty$. As a result $\big(1,\tfrac{1}{2}\big)$ is a global (and not just local) minimum for $f$ in the first quadrant. Hence $f(x,y)\ge f\big(1,\tfrac{1}{2}\big)=6$ for all $x,y>0$.