Solutions To Selected Problems Chapter 2 Chapter 4

Chapter 3

Exercises 3.1

1.

Answers will vary.
3.

topographical
5.

surface
7.

domain: $\mathbb{R}^{2}$

range: $z\geq 2$
9.

domain: $\mathbb{R}^{2}$

range: $\mathbb{R}$
11.

domain: $\mathbb{R}^{2}$

range: $0<z\leq 1$
13.

domain: $\{(x,y)\,|\,x^{2}+y^{2}\leq 9\}$ , i.e., the domain is the circle and interior of a circle centered at the origin with radius 3.

range: $0\leq z\leq 3$
15.

Level curves are lines $y=(3/2)x-c/2$ .
17.

Level curves are parabolas $x=y^{2}+c$ .
19.

Level curves are circles, centered at $(1/c,-1/c)$ with radius $\sqrt{2/c^{2}-1}$ . When $c=0$ , the level curve is the line $y=x$ .
21.

Level curves are ellipses of the form $\frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}/4}=1$ , i.e., $a=c$ and $b=c/2$ .
23.

domain: $x+2y-4z\neq 0$ ; the set of points in $\mathbb{R}^{3}$ NOT in the domain form a plane through the origin.

range: $\mathbb{R}$
25.

domain: $z\geq x^{2}-y^{2}$ ; the set of points in $\mathbb{R}^{3}$ above (and including) the hyperbolic paraboloid $z=x^{2}-y^{2}$ .

range: $[0,\infty)$
27.

The level surfaces are spheres, centered at the origin, with radius $\sqrt{c}$ .
29.

The level surfaces are paraboloids of the form $z=\frac{x^{2}}{c}+\frac{y^{2}}{c}$ ; the larger $c$ , the “wider” the paraboloid.
31.

The level curves for each surface are similar; for $z=\sqrt{x^{2}+4y^{2}}$ the level curves are ellipses of the form $\frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}/4}=1$ , i.e., $a=c$ and $b=c/2$ ; whereas for $z=x^{2}+4y^{2}$ the level curves are ellipses of the form $\frac{x^{2}}{c}+\frac{y^{2}}{c/4}=1$ , i.e., $a=\sqrt{c}$ and $b=\sqrt{c}/2$ . The first set of ellipses are spaced evenly apart, meaning the function grows at a constant rate; the second set of ellipses are more closely spaced together as $c$ grows, meaning the function grows faster and faster as $c$ increases.

The function $z=\sqrt{x^{2}+4y^{2}}$ can be rewritten as $z^{2}=x^{2}+4y^{2}$ , an elliptic cone; the function $z=x^{2}+4y^{2}$ is a paraboloid, each matching the description above.

Exercises 3.2

1.

Answers will vary.
3.

Answers will vary.

One possible answer: $\{(x,y)|x^{2}+y^{2}\leq 1\}$
5.

Answers will vary.

One possible answer: $\{(x,y)|x^{2}+y^{2}<1\}$
7.
[-2]
1. (a)
  
  Answers will vary.
  
  interior point: $(1,3)$
  
  boundary point: $(3,3)$
2. (b)
  
  $S$ is a closed set
3. (c)
  
  $S$ is bounded
9.
[-2]
1. (a)
  
  Answers will vary.
  
  interior point: none
  
  boundary point: $(0,-1)$
2. (b)
  
  $S$ is a closed set, consisting only of boundary points
3. (c)
  
  $S$ is bounded
11.
[-2]
1. (a)
  
  $D=\left\{(x,y)\,|\,9-x^{2}-y^{2}\geq 0\right\}$ .
2. (b)
  
  $D$ is a closed set.
3. (c)
  
  $D$ is bounded.
13.
[-2]
1. (a)
  
  $D=\left\{(x,y)\,|\,y>x^{2}\right\}$ .
2. (b)
  
  $D$ is an open set.
3. (c)
  
  $D$ is unbounded.
15.
[-2]
1. (a)
  
  $D=\mathbb{R}^{2}$ .
2. (b)
  
  $D$ is an open set.
3. (c)
  
  $D$ is unbounded.
17.
[-2]
1. (a)
  
  $D=\{(x,y)\,|\,x^{2}+y^{2}<1\}$ . This is the open disk of radius 1 centered at the origin.
2. (b)
  
  $D$ is an open set.
3. (c)
  
  $D$ is bounded.
19.
[-2]
1. (a)
  
  Along $y=0$ , the limit is 1.
2. (b)
  
  Along $x=0$ , the limit is $-1$ .
Since the above limits are not equal, the limit does not exist.
21.
[-2]
1. (a)
  
  Along $y=mx$ , the limit is $0$ .
2. (b)
  
  Along $x=0$ , the limit is $-1$ .
Since the above limits are not equal, the limit does not exist.
23.
[-2]
1. (a)
  
  Along $y=2$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$ $\displaystyle=\lim_{x\to 1}\frac{x-1}{x^{2}-1}$
  
  $\displaystyle=\lim_{x\to 1}\frac{1}{x+1}$
  
  $\displaystyle=1/2.$
2. (b)
  
  Along $y=x+1$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$ $\displaystyle=\lim_{x\to 1}\frac{2(x-1)}{x^{2}-1}$
  
  $\displaystyle=\lim_{x\to 1}\frac{2}{x+1}$
  
  $\displaystyle=1.$
Since the limits along the lines $y=2$ and $y=x+1$ differ, the overall limit does not exist.
25.

Hint: Consider $-\left\lvert\sin(xy)\right\rvert\leq\sin(xy)\cos\left(\frac{1}{x^{2}+y^{2}}% \right)\leq\left\lvert\sin(xy)\right\rvert$ .
27.

Hint: $x^{4}-9y^{4}=(x^{2}+3y^{2})(x^{2}-3y^{2})$ .

Exercises 3.3

1.

A constant is a number that is added or subtracted in an expression; a coefficient is a number that is being multiplied by a nonconstant function.
3.

$f_{x}$
5.

$f_{x}=2xy-1$ , $f_{y}=x^{2}+2$

$f_{x}(1,2)=3$ , $f_{y}(1,2)=3$
7.

$f_{x}=-\sin x\sin y$ , $f_{y}=\cos x\cos y$

$f_{x}(\pi/3,\pi/3)=-3/4$ , $f_{y}(\pi/3,\pi/3)=1/4$
9.

$f_{x}=2xy+6x$ , $f_{y}=x^{2}+4$

$f_{xx}=2y+6$ , $f_{yy}=0$

$f_{xy}=2x$ , $f_{yx}=2x$
11.

$f_{x}=1/y$ , $f_{y}=-x/y^{2}$

$f_{xx}=0$ , $f_{yy}=2x/y^{3}$

$f_{xy}=-1/y^{2}$ , $f_{yx}=-1/y^{2}$
13.

$f_{x}=2xe^{x^{2}+y^{2}}$ , $f_{y}=2ye^{x^{2}+y^{2}}$

$f_{xx}=2e^{x^{2}+y^{2}}+4x^{2}e^{x^{2}+y^{2}}$ , $f_{yy}=2e^{x^{2}+y^{2}}+4y^{2}e^{x^{2}+y^{2}}$

$f_{xy}=4xye^{x^{2}+y^{2}}$ , $f_{yx}=4xye^{x^{2}+y^{2}}$
15.

$f_{x}=\cos x\cos y$ , $f_{y}=-\sin x\sin y$

$f_{xx}=-\sin x\cos y$ , $f_{yy}=-\sin x\cos y$

$f_{xy}=-\sin y\cos x$ , $f_{yx}=-\sin y\cos x$
17.

$f_{x}=-5y^{3}\sin(5xy^{3})$ , $f_{y}=-15xy^{2}\sin(5xy^{3})$

$f_{xx}=-25y^{6}\cos(5xy^{3})$ , $f_{yy}=-225x^{2}y^{4}\cos(5xy^{3})-30xy\sin(5xy^{3})$

$f_{xy}=-75xy^{5}\cos(5xy^{3})-15y^{2}\sin(5xy^{3})$ , $f_{yx}=-75xy^{5}\cos(5xy^{3})-15y^{2}\sin(5xy^{3})$
19.

$f_{x}=\frac{2y^{2}}{\sqrt{4xy^{2}+1}}$ , $f_{y}=\frac{4xy}{\sqrt{4xy^{2}+1}}$

$f_{xx}=-\frac{4y^{4}}{\sqrt{4xy^{2}+1}^{3}}$ , $f_{yy}=-\frac{16x^{2}y^{2}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4x}{\sqrt{4xy^{2}+1}}$

$f_{xy}=-\frac{8xy^{3}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4y}{\sqrt{4xy^{2}+1}}$ , $f_{yx}=-\frac{8xy^{3}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4y}{\sqrt{4xy^{2}+1}}$
21.

$f_{x}=-\frac{2x}{(x^{2}+y^{2}+1)^{2}}$ , $f_{y}=-\frac{2y}{(x^{2}+y^{2}+1)^{2}}$

$f_{xx}=\frac{8x^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{2}{(x^{2}+y^{2}+1)^{2}}$ , $f_{yy}=\frac{8y^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{2}{(x^{2}+y^{2}+1)^{2}}$

$f_{xy}=\frac{8xy}{(x^{2}+y^{2}+1)^{3}}$ , $f_{yx}=\frac{8xy}{(x^{2}+y^{2}+1)^{3}}$
23.

$f_{x}=6x$ , $f_{y}=0$

$f_{xx}=6$ , $f_{yy}=0$

$f_{xy}=0$ , $f_{yx}=0$
25.

$f_{x}=\frac{1}{4xy}$ , $f_{y}=-\frac{\ln x}{4y^{2}}$

$f_{xx}=-\frac{1}{4x^{2}y}$ , $f_{yy}=\frac{\ln x}{2y^{3}}$

$f_{xy}=-\frac{1}{4xy^{2}}$ , $f_{yx}=-\frac{1}{4xy^{2}}$
27.

$f(x,y)=x\sin y+x+C$ , where $C$ is any constant.
29.

$f(x,y)=3x^{2}y-4xy^{2}+2y+C$ , where $C$ is any constant.
31.

$f_{x}=2xe^{2y-3z}$ , $f_{y}=2x^{2}e^{2y-3z}$ , $f_{z}=-3x^{2}e^{2y-3z}$

$f_{yz}=-6x^{2}e^{2y-3z}$ , $f_{zy}=-6x^{2}e^{2y-3z}$
33.

$f_{x}=\frac{3}{7y^{2}z}$ , $f_{y}=-\frac{6x}{7y^{3}z}$ , $f_{z}=-\frac{3x}{7y^{2}z^{2}}$

$f_{yz}=\frac{6x}{7y^{3}z^{2}}$ , $f_{zy}=\frac{6x}{7y^{3}z^{2}}$

Exercises 3.4

1.

T
3.

T
5.

$dz=(\sin y+2x)dx+(x\cos y)dy$
7.

$dz=5dx-7dy$
9.

$dz=\frac{x}{\sqrt{x^{2}+y}}dx+\frac{1}{2\sqrt{x^{2}+y}}dy$ , with $dx=-0.05$ and $dy=.1$ . At $(3,7)$ , $dz=3/4(-0.05)+1/8(.1)=-0.025$ , so $f(2.95,7.1)\approx-0.025+4=3.975$ .
11.

$dz=(2xy-y^{2})dx+(x^{2}-2xy)dy$ , with $dx=0.04$ and $dy=0.06$ . At $(2,3)$ , $dz=3(0.04)+(-8)(0.06)=-0.36$ , so $f(2.04,3.06)\approx-0.36-6=-6.36$ .
13.

The total differential of volume is $dV=4\pi dr+\pi dh$ . The coefficient of $d r$ is greater than the coefficient of $d h$ , so the volume is more sensitive to changes in the radius.
15.

Using trigonometry, $\ell=x\tan\theta$ , so $d\ell=\tan\theta dx+x\sec^{2}\theta d\theta$ . With $\theta=85^{\circ}$ and $x=30$ , we have $d\ell=11.43dx+3949.38d\theta$ . The measured length of the wall is much more sensitive to errors in $\theta$ than in $x$ . While it can be difficult to compare sensitivities between measuring feet and measuring degrees (it is somewhat like “comparing apples to oranges”), here the coefficients are so different that the result is clear: a small error in degree has a much greater impact than a small error in distance.
17.

$dw=2xyz^{3}\,dx+x^{2}z^{3}\,dy+3x^{2}yz^{2}\,dz$
19.

$dx=0.05$ , $dy=-0.1$ . $dz=9(.05)+(-2)(-0.1)=0.65$ . So $f(3.05,0.9)\approx 7+0.65=7.65$ .
21.

$dx=0.5$ , $dy=0.1$ , $dz=-0.2$ .

$dw=2(0.5)+(-3)(0.1)+3.7(-0.2)=-0.04$ , so $f(2.5,4.1,4.8)\approx-1-0.04=-1.04$ .
23.

Everywhere except the origin.

Exercises 3.5

1.

Because the parametric equations describe a level curve, $z$ is constant for all $t$ . Therefore $\frac{dz}{dt}=0$ .
3.

$\frac{dx}{dt}$ , and $\frac{\partial f}{\partial y}$
5.

F
7.
[-2]
1. (a)
  
  $\frac{dz}{dt}=3(2t)+4(2)=6t+8$ .
2. (b)
  
  At $t=1$ , $\frac{dz}{dt}=14$ .
9.
[-2]
1. (a)
  
  $\frac{dz}{dt}=5(-2\sin t)+2(\cos t)=-10\sin t+2\cos t$
2. (b)
  
  At $t=\pi/4$ , $\frac{dz}{dt}=-4\sqrt{2}$ .
11.
[-2]
1. (a)
  
  $\displaystyle\frac{dz}{dt}=2x(\cos t)+4y(3\cos t)$ .
2. (b)
  
  At $t=\pi/4$ , $x=\sqrt{2}/2$ , $y=3\sqrt{2}/2$ , and $\frac{dz}{dt}=19$ .
13.

$t=-4/3$ ; this corresponds to a minimum
15.

$t=\tan^{-1}(1/5)+n\pi$ , where $n$ is an integer
17.
We find that $\frac{dz}{dt}=38\cos t\sin t.$ Thus $\frac{dz}{dt}=0$ when $t=\pi n$ or $\pi n+\pi/2$ , where $n$ is any integer.
19.
[-2]
1. (a)
  
  $\frac{\partial z}{\partial s}=2xy(1)+x^{2}(2)=2xy+2x^{2}$ ;
  
  $\frac{\partial z}{\partial t}=2xy(-1)+x^{2}(4)=-2xy+4x^{2}$
2. (b)
  
  With $s=1$ , $t=0$ , $x=1$ and $y=2$ . Thus $\frac{\partial z}{\partial s}=6$ and $\frac{\partial z}{\partial t}=0$
21.
[-2]
1. (a)
  
  $\frac{\partial z}{\partial s}=2x(\cos t)+2y(\sin t)=2x\cos t+2y\sin t$ ;
  
  $\frac{\partial z}{\partial t}=2x(-s\sin t)+2y(s\cos t)=-2xs\sin t+2ys\cos t$
2. (b)
  
  With $s=2$ , $t=\pi/4$ , $x=\sqrt{2}$ and $y=\sqrt{2}$ . Thus $\frac{\partial z}{\partial s}=4$ and $\frac{\partial z}{\partial t}=0$
23.

$f_{x}=2x\tan y$ , $f_{y}=x^{2}\sec^{2}y$ ;

$\displaystyle\frac{dy}{dx}=-\frac{2\tan y}{x\sec^{2}y}$
25.

$\displaystyle f_{x}=\frac{(x+y^{2})(2x)-(x^{2}+y)(1)}{(x+y^{2})^{2}}$ ,

$\displaystyle f_{y}=\frac{(x+y^{2})(1)-(x^{2}+y)(2y)}{(x+y^{2})^{2}}$ ;

$\displaystyle\frac{dy}{dx}=-\frac{2x(x+y^{2})-(x^{2}+y)}{x+y^{2}-2y(x^{2}+y)}$
27.

$\frac{dz}{dt}=2(4)+1(-5)=3$ .
29.

$\frac{\partial z}{\partial s}=-4(5)+9(-2)=-38$ ,

$\frac{\partial z}{\partial t}=-4(7)+9(6)=26$ .
31.

$z_{\theta}=-f_{x}(x,y)r\sin\theta+f_{y}(x,y)r\cos\theta$
33.

It is increasing at $104\pi/3$ cm³/sec

Exercises 3.6

1.

A partial derivative is essentially a special case of a directional derivative; it is the directional derivative in the direction of $x$ or $y$ , i.e., $\left\langle 1,0\right\rangle$ or $\left\langle 0,1\right\rangle$ .
3.

$\vec{u}=\left\langle 0,1\right\rangle$
5.

maximal, or greatest
7.

$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$
9.

$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$
11.

$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$
13.
$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$ ; $\nabla f(2,1)=\left\langle-2,2\right\rangle$ . Be sure to change all directions to unit vectors. (a) $2/5$ ( $\vec{u}=\left\langle 3/5,4/5\right\rangle$ ) (b) $-2/\sqrt{5}$ ( $\vec{u}=\left\langle-1/\sqrt{5},-2/\sqrt{5}\right\rangle$ )
15.
$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$ ; $\nabla f(1,1)=\left\langle-2/9,-2/9\right\rangle$ . Be sure to change all directions to unit vectors. (a) 0 ( $\vec{u}=\left\langle 1/\sqrt{2},-1/\sqrt{2}\right\rangle$ ) (b) $2\sqrt{2}/9$ ( $\vec{u}=\left\langle-1/\sqrt{2},-1/\sqrt{2}\right\rangle$ )
17.
$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$ ; $\nabla f(4,1)=\left\langle 0,0\right\rangle$ . (a) 0 (b) 0
19.
$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$ (a) $\nabla f(2,1)=\left\langle-2,2\right\rangle$ (b) $\left\lVert\nabla f(2,1)\right\rVert=\left\lVert\left\langle-2,2\right\rangle% \right\rVert=\sqrt{8}$ (c) $\left\langle 2,-2\right\rangle$ (d) $\left\langle 1/\sqrt{2},1/\sqrt{2}\right\rangle$
21.
$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$ (a) $\nabla f(1,1)=\left\langle-2/9,-2/9\right\rangle$ . (b) $\left\lVert\nabla f(1,1)\right\rVert=\left\lVert\left\langle-2/9,-2/9\right% \rangle\right\rVert=2\sqrt{2}/9$ (c) $\left\langle 2/9,2/9\right\rangle$ (d) $\left\langle 1/\sqrt{2},-1/\sqrt{2}\right\rangle$
23.
$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$ (a) $\nabla f(4,1)=\left\langle 0,0\right\rangle$ (b) 0 (c) $\left\langle 0,0\right\rangle$ (d) All directions give a directional derivative of 0.
25.
(a) $\nabla F(x,y,z)=\left\langle 6xz^{3}+4y,4x,9x^{2}z^{2}-6z\right\rangle$ (b) $113/\sqrt{3}$
27.
(a) $\nabla F(x,y,z)=\left\langle 2xy^{2},2y(x^{2}-z^{2}),-2y^{2}z\right\rangle$ (b) $0$
29.

In the direction $\left\langle 7,8,-4\right\rangle$ with maximal value $\sqrt{129}$ .

Exercises 3.7

1.

Answers will vary. The displacement of the vector is one unit in the $x$ -direction and 3 units in the $z$ -direction, with no change in $y$ . Thus along a line parallel to $\vec{v}$ , the change in $z$ is 3 times the change in $x$ – i.e., a “slope” of 3. Specifically, the line in the $x$ - $z$ plane parallel to $z$ has a slope of 3.
3.

T
5.
[-2]
1. (a)
  
  $\ell_{x}(t)=\begin{cases}x=2+t\\ y=3\\ z=-48-12t\end{cases}$
2. (b)
  
  $\ell_{y}(t)=\begin{cases}x=2\\ y=3+t\\ z=-48-40t\end{cases}$
3. (c)
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=2+t/\sqrt{10}\\ y=3+3t/\sqrt{10}\\ z=-48-66\sqrt{2/5}t\end{cases}$
7.
[-2]
1. (a)
  
  $\ell_{x}(t)=\begin{cases}x=4+t\\ y=2\\ z=2+3t\end{cases}$
2. (b)
  
  $\ell_{y}(t)=\begin{cases}x=4\\ y=2+t\\ z=2-5t\end{cases}$
3. (c)
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=4+t/\sqrt{2}\\ y=2+t/\sqrt{2}\\ z=2-\sqrt{2}t\end{cases}$
9.

$\ell_{\vec{n}}(t)=\begin{cases}x=2-12t\\ y=3-40t\\ z=-48-t\end{cases}$
11.

$\ell_{\vec{n}}(t)=\begin{cases}x=4+3t\\ y=2-5t\\ z=2-t\end{cases}$
13.

$(1.425,1.085,-48.078)$ , $(2.575,4.915,-47.952)$
15.

$(5.014,0.31,1.662)$ and $(2.986,3.690,2.338)$
17.

$-12(x-2)-40(y-3)-(z+48)=0$
19.

$3(x-4)-5(y-2)-(z-2)=0$ (Note that this tangent plane is the same as the original function, a plane.)
21.
$\nabla F=\left\langle x/4,y/2,z/8\right\rangle$ ; at $P$ , $\nabla F=\left\langle 1/4,\sqrt{2}/2,\sqrt{6}/8\right\rangle$ (a) $\ell_{\vec{n}}(t)=\begin{cases}x=1+t/4\\ y=\sqrt{2}+\sqrt{2}t/2\\ z=\sqrt{6}+\sqrt{6}t/8\end{cases}$ (b) $\frac{1}{4}(x-1)+\frac{\sqrt{2}}{2}(y-\sqrt{2})+\frac{\sqrt{6}}{8}(z-\sqrt{6})=0$ .
23.
$\nabla F=\left\langle y^{2}-z^{2},2xy,-2xz\right\rangle$ ; at $P$ , $\nabla F=\left\langle 0,4,4\right\rangle$ (a) $\ell_{\vec{n}}(t)=\begin{cases}x=2\\ y=1+4t\\ z=-1+4t\end{cases}$ (b) $4(y-1)+4(z+1)=0$ .

Exercises 3.8

1.

F; it is the “other way around.”
3.

T
5.

One critical point at $(-4,2)$ ; $f_{xx}=1$ and $D=4$ , so this point corresponds to a relative minimum.
7.

One critical point at $(6,-3)$ ; $D=-4$ , so this point corresponds to a saddle point.
9.

Two critical points: at $(0,-1)$ ; $f_{xx}=2$ and $D=-12$ , so this point corresponds to a saddle point;

at $(0,1)$ , $f_{xx}=2$ and $D=12$ , so this corresponds to a relative minimum.
11.

Critical points when $x$ or $y$ are $0$ . $D=-12x^{2}y^{2}$ , so the test is inconclusive. (Some elementary thought shows that these are absolute minima.)
13.

One critical point: $f_{x}=0$ when $x=3$ ; $f_{y}=0$ when $y=0$ , so one critical point at $(3,0)$ , which is a relative maximum, where $f_{xx}=\frac{y^{2}-16}{(16-(x-3)^{2}-y^{2})^{3/2}}$ and $D=\frac{16}{(16-(x-3)^{2}-y^{2})^{2}}$ .

Both $f_{x}$ and $f_{y}$ are undefined along the circle $(x-3)^{2}+y^{2}=16$ ; at any point along this curve, $f(x,y)=0$ , the absolute minimum of the function.
15.

rel. max at $(0,0)$ ; rel. min at $(2,0)$ ; saddle points at $(1,\pm 1)$ .
17.

saddle points at $(1,2/3)$ and $(-1,-4/3)$ .
19.

The triangle is bound by the lines $y=-1$ , $y=2x+1$ and $y=-2x+1$ .

Along $y=-1$ , there is a critical point at $(0,-1)$ .

Along $y=2x+1$ , there is a critical point at $(-3/5,-1/5)$ .

Along $y=-2x+1$ , there is a critical point at $(3/5,-1/5)$ .

The function $f$ has one critical point, irrespective of the constraint, at $(0,-1/2)$ .

Checking the value of $f$ at these four points, along with the three vertices of the triangle, we find the absolute maximum is at $(0,1,3)$ and the absolute minimum is at $(0,-1/2,3/4)$ .
21.

The region has no “corners” or “vertices,” just a smooth edge.

To find critical points along the circle $x^{2}+y^{2}=4$ , we solve for $y^{2}$ : $y^{2}=4-x^{2}$ . We can go further and state $y=\pm\sqrt{4-x^{2}}$ .

We can rewrite $f$ as $f(x)=x^{2}+2x+(4-x^{2})+2\sqrt{4-x^{2}}=2x+4+2\sqrt{4-x^{2}}$ . (We will return and use $-\sqrt{4-x^{2}}$ later.) Solving $f\,^{\prime}(x)=0$ , we get $x=\sqrt{2}\Rightarrow y=\sqrt{2}$ . $f\,^{\prime}(x)$ is also undefined at $x=\pm 2$ , where $y=0$ .

Using $y=-\sqrt{4-x^{2}}$ , we rewrite $f(x,y)$ as $f(x)=2x+4-2\sqrt{4-x^{2}}$ . Solving $f\,^{\prime}(x)=0$ , we get $x=-\sqrt{2},\ y=-\sqrt{2}$ .

The function $f$ itself has a critical point at $(-1,-1)$ .

Checking the value of $f$ at $(-1,-1)$ , $(\sqrt{2},\sqrt{2})$ , $(-\sqrt{2},-\sqrt{2})$ , $(2,0)$ and $(-2,0)$ , we find the absolute maximum is at $(\sqrt{2},\sqrt{2},4+4\sqrt{2})$ and the absolute minimum is at $(-1,-1,-2)$ .
23.

abs max is $(1,2,6)$ , abs min is $(3,0,-6)$ .
25.

$10\times 10\times 5$

Exercises 3.9

1.

$\pm 2\sqrt{5}$ at $(\pm 4/\sqrt{5},\pm 2/\sqrt{5})$
3.

$(\pm 20/\sqrt{13},\pm 30/\sqrt{13})$
5.

$(2/\sqrt{3})^{3}=8/3\sqrt{3}$
7.

Length 130/3, height and width 65/3.
9.

$(0,\pm 1,0)$
11.

$(2,1,2)$
13.

Max: 5 at $\pm(2,2)$ , min: $-9/2$ at $\pm(3/\sqrt{2},-3/\sqrt{2})$ .
15.

$8abc/3\sqrt{3}$
17.

$f(x,x^{2}-1)\to\infty$ as $x\to\infty$ . Minimum is $(-3/4,-7/16,-155/128)$ .

	$\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$	$\displaystyle=\lim_{x\to 1}\frac{x-1}{x^{2}-1}$
		$\displaystyle=\lim_{x\to 1}\frac{1}{x+1}$
		$\displaystyle=1/2.$

	$\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$	$\displaystyle=\lim_{x\to 1}\frac{2(x-1)}{x^{2}-1}$
		$\displaystyle=\lim_{x\to 1}\frac{2}{x+1}$
		$\displaystyle=1.$