数学分析简明教程答案

第十八章§1（1）f(x,y)(xy1)2（2）f(x,y)3axyx3y3(a0)1 x ya1 x ya b2f(x,y)e2x(xy22y)f(x,y)sinxcosycos(x

（0x,y 2x2yf(x,y)x2yfx(x,y)2(xy1)0f解（1）由f

y(x,y)2(xy1)0

yx1fxx2fxy2fyy2D0fyxf(xy)0yx1f(xy)0yx1f(xy也是最小值0（2）由

(x,y)3ay3x20 解出稳定点为(0,0),(aafy(x,y)3ax3y20在点(0,0)a11

fxx(0,0)0，

fxy(0,0)3aa22fyy(0,00D

29a20故(0,0)不是极值点．在点(aaa11

fxx(a,a)6a，

fxy(a,a)3a，a22fyy(a,a)6aD

227a20，

6a0f(x,y在(aaf(aa)a3

0,

,

0D

240f(x,y

a

b

3ab23233333（4）令

(x,y)e2x(12x4y2y2)01 解得稳定点1

fy(x,y)2e2x(1y)0 a11

fxx(2,1)2e0,a22fyy(2,1)2e0,a12fxy(2,1)0D

24e20f(xy在1,1

e121fx(x,y)cosxsin(xy)0 f（5）令f33

y(x,y)sinysin(xy)0

解得稳定点为 )3a11

fxx(

,)6

0,a22

fyy(

,)6

30,a12

fxy(

,) 3 3Da

a290，故f(x,y)在(,)取极大值为 311

3 f(x,y)

0x2x2yx2y

x2y21x2yx2yfx2yx2y

0由于f(xy)x2y21上的点取值0，而f(xy)0，故f(x,y)在圆周1x2y21上的点取极小值也是最小值0P(0,01x2yf(x,x2y

1)21

f(0,0)，

2x2x2 yax2bxc(xyi1,2,nab n解f(abc)(ax2bxcy)2abcn 并令它们等于0f

(ax2

cy)x20ii iif

(ax2

cy

0 ii iif

(ax2

icy)0 iii iiabcxixi

xxxxx,i ia

nx3

x2

nx

nxy

n

n

iiiax2bxcnyiii

A1(x1y1A2(x2y2,An(xnyn，n个点距离的平方和最小．解P(xynnf(x,y)[(xx)2(yy)2n

f

(xx)0，f

i(yy)0ii i得稳定点1n

x,1ini

yi)(x,y点即为最小值点．因而点(x,yn（1）f(x,y)x2y2,D(x,y)x2y2（2）f(x,y)x2xyy2,D(x,y)xy（3）f(xyz)axbycz)e(x2y2z2)a2b2c20DR3fx(x,y)2x0f解（1）令f

y(x,y)2y0

D内的唯一稳定点(0,0)a11

fxx(0,0)20,

fyy(0,0)20,

fxy(0,0)0D

240(0,0)点，f(xyx2y24f(x,y)x2y2x2(4x2)2(x22)(x),x2令(x4x0x0，且(x40，故(xx0这就是函数f(xy)的最小值，其值为

f(0,2)4x2y0f(2,0)4，即函数在(2,0取最大值4fx(x,y)2xy0,f（2）令f

y(x,y)2yx0

解得唯一的稳定点(0,0)a11

fxx(0,0)20,

fyy(0,0)20,

fxy(0,0)D

250(0,0)

f(x,y)的极小值点，也是最小值点，最小值为f(0,0)0xy1y1x0x1f(x,y)(13x)23x2 x0x1f(1,0)f(0,1)1x1f(11)1 2 y1x1x0f(x,y)x2x1x0x1f(1,0)f(0,1)1x1f(11)3 2 y1x0xf(x,y)x2x1x0x1f(1,0)f(0,1)1x1f(1,1)12Df(xyf(0,0)0，取最大值（3）

f(x,y,z)[a2x(axbycz)]e(x2y2z2)0xyz22yz

(x,y,z)[b2y(axbycz)]e(xyz)0,(x,y,z)[c2z(axbycz)]e(x2y2z2)0,

2(a2(a2b2c2

2(a22(a2b2c22(a2b2c22(a2b2c2

)，2(a2b22(a2b2c22(a2b2c2a2b2c2Hessef(xa2b2c2a2b2ca2b2c2

2，最小值为f(P2) e2f(xyAx22BxyCy22Dx2EyFR2A0,B2ACf(xy)xy11在0x,y fx(x,y)2Ax2By2D0f证明（1）令f

y(x,y)2Bx2Cy2E0B

BACB20CP0(x0y0a11

fxx(P0)2A0,

fyy(P0)2C,

fxy(P0)2BD

24B24AC0f(xyP0f(x,yR20（2）f2 (x,y)yf2x1

0

fy(x,y)xy

0a11

fxx(P0)20,

fyy(P0)20,

fxy(P0)1且Da a

30f(xyxy11P(1,111

f(xy在0x,yF(x,yzF(x0,y0,z0)0，Fz(x0,y0,z0)0F(xyz)0zf(xy在(x0y0由x2y2z22x2y4z10z

f(x,y解取极值的必要条件为在(x0y0

f(x0,y0)Fx(x0,y0,z0)0

(x0,y0

Fz(x0,y0,z0

f(x,y

F(x,y,z 0

00(x,y

Fz(x0,y0,z0 0Fx(x0y0z0)0Fy(x0y0z0)0为在(x0y0取得极值的必要条件．F(x,yz)0x,y二次求导，有FFz0 z Fy

0

z

FxxFxyxFzxxFzz(x

Fzx

0

z

FxyFxzyFzyxFzzxyFzxy0 z

FyyFyzyFzyyFzz(y并利用在(xyz0z0

Fzy

0 F(x

Fxy(x0,y0,z0 0(x0,y0

Fz

,

,

(x0,y0

Fz

,

,z0Fyy(x0,y0,(x0,y0

Fz(x0,y0,z0因此在(x0y0z

f(x,y)(Fxx(x0,y0,z0))(Fyy(x0,y0,z0))(Fxy(x0,y0,z0))2Fz(x0,y0,z0

Fz(x0,y0,z0

Fz(x0,y0,z000

,

,

)Fyy(x0,

,z0)F2

,

,z0

0z0z0

,y0,z0F(xyz

(x,y,

)F2(x,y,

)0Fxx(x0y0z0)0（

Fz(x0,y0,z0Fyy(x0y0z0)0Fxx(x0y0z0)0（或Fyy(x0y0z0)0Fz(x0,y0,z0

Fz(x0,y0,z0

Fz(x0,y0,z0

,

,

,

,

)F2

,

,

00取极值，为00F(xyz)x2y2z22x2y4z10Fx(x,y,z)2x20F (x,y,z)2y20FFxx2,Fxy0,Fyy2因为DF F

40（在P与 ），且F(P)(2z4)

8xx

2Fz(P2)2z4|P82由于Fxx(P1)20P(1,1,6取极大值6；而Fxx(P2)20F(P

F(P （1）(xy)2(yz)2(zx)23（2）z2xyzx2xy290解（1）F(xyz)(xy)2yz)2zx)23Fx(x,y,z)2(xy)2(zx)0F (x,y,z)2(xy)2(zy)0F31F(x,y,z)0311

12

12

12

Fxx4,Fxy2,Fyy4D

212012Fz(P1)[2(yz)2(zx)]|P4，Fz(P2)[2(yz)2(zx)]12

4因此，Fxx(P1)10(11取极小值3；Fxx(P2)10(1,11取极大值 12

2

Fz(P2 （2）F(xyz)z2xyzx2xy29F(x,y,z)yz2xy20Fy(x,y,z)xz2xy0F(x,y,z)0

Fxx2,Fxyz2y,Fyy2xP1(0,0,3Fxx(P12,Fxy(P13,Fyy(P10D90P2(0,0,3Fxx(P22,Fxy(P23,Fyy(P20，D90，故函数P20,0,3)不取极值；P3(0,3,3Fxx(P32Fxy(P33Fyy(P30，D90P4(0,3,3Fxx(P42,Fxy(P43,Fyy(P40，D90，故函数P40,3,3)也不取极值；

Fxx(P52,Fxy(P50,Fyy(P52，D4025Fz(P5)(2zxy)]|P25

Fxx(P5)

202 2

取极小值

Fxx(P62,Fxy(P60,Fyy(P62，D406Fz(P62zxy|P6

2Fxx(P6)2

20

Fz(P6 2取极大值 2在已知周长为2p的一切三角形中，求出面积为最大的三角形．解x,y，则另一边长为2pxyp(px)(p(px)(py)(p(2pxp(px)(py)(xyss2s2p(px)(py)(xyp)F(x,y)令Fx(x,y)p(py)(2p2xy)0F (x,y,z)p(px)(2px2y)0F解得稳定点2p3

2p．由于驻点唯一，实际问题又有最大值，故最大值点为2p

2p)3z2p，即在已知周长为2p3的倾角x（见下图．解s1

s xsin(242x242x2xcos)2sx(244x2xcos)sin0s 24xcosx2cos22x2cos0s 解得稳定点为

3

3倾角x8cm3§2

xyx2y21；

x2y2xy10

x2y2zx2y2z211xy2 fxyzx2y2z21xyz0

ax2by22hxyx2y21；

x2y2z2，若(x2y2z22a2x2b2y2c2z2lxmynz0解（1）LagrangeL(xy)xy(x2y21Lx12x0Ly12y0x2y21由前两式得xy，代入最后一式解得稳定点P(2 2)与P 2 F(xy)x2y21F(xy)2yP,P 等于0x2y21PPyy(x g(x)xy(xdyxdg1dy

1

，

yx y

1ygx

2d2d2

222

112

0因此g(x)在x 处2222222

xy在P(2 2)处取得条件极大值f(P) 2 2函数在x22d2212220g(x)22x

处取极小值，这等价于2

fxyP2

22

处取得条件极小值22f(P2) 2LagrangeL(xy)x2y2(xy1Lx2x0L 2y0Lxy2

xy1011xy2P0211

1)2F(xy)xy1Fy(xy)10xy10yy(xg(x)x2y2xdy1dg2x2ydy

2x2y

22

401所以函数g在x 取极小值，这等价于1121f(P0)2

x2y2

1

处取得条件极小值作Lagrange函数L(x,y,z)x2y2z ，Lx12x0L 22y0LL 22z0Lyz2xP(1,22P(12,21 3 3 F(xyz)x2y2z21F(xyz)2zPP均不等于0，故方程 x2y2z21PPzz(xy g(x,y)x2y2z(x,y) xyyz

2x12x1，2z2x，

z，y22y2z2x2g

x2(x2z2 2g

1g在(,13

z2g3a

z2z2yz

z z2(y2z2z3

2 32

2180且 150，因此g(x,y)在(1,2)处取极大值，这等价

f(xyz)11

2 ,)f(,3

2,3

)3g在(13

3

12 2

21806且 150，故g(x,y)在(1,2)处取极小值，这等价于f(x,y,z)在P(

2

31

3,3,处取得条件极小值f( 3

)33 作Lagrange函数L(x,y) x

(xy2)yLL L

1λ0x10 y由前两式得

xy201x2y2yxxy1（yF(x,y)xy2Fy(xy)10xy20yy(xg(x,y)1x

，由约束条件 1，所以dg1 dy

1 dg

dy22 y2(x)

x2，

y3 y

40g(xx1f(xy在(1,1LagrangeL(xyz)xyz1Lxyz21x20

LL

xz21y20

xy21z20x2y2z21xyzxy21xz21yz21xyxy

21xz21yz2166xy1，z266或6xz6

6，y26或6yz6

6，x26f(xyzxyz,x2y2z21xyz0x,yz166 66

1.

616616

2626

)16若记F(xyz)x2y2z21G(xyz)xyz16(F,G)2

2z2(yz)(y,

x2y2z2P20，故方程组xyz

P2yy(x),zz(xg(x)xy(x)z(xdyyz

x ，z

y zdgyzxdy

zxydz

yz2y2zx2zxz2xy2x2，zd2g4(xz3yz3y3zxy3x2y2)2(y4z4)12xyz(y1

66,166

62x6

166

(zd2d26

660gx6f(xyzP1

f

166166

) 62626

1616666666

x

d2d26

60g16x f(xyzP216f

1616

1616

) 266266 6 1,2, 6 1,2,与2,1,6

2,

与

2,1

1两点取条件极小值6 LagrangeL(xy)ax2by22hxy(x2y21Lx2ax2hy2x0Ly2by2hx2y0x2y21x2y21xy不全为0xy的线性方程组的系数矩阵必等于0即aAh

2(ab)abh20 当(ab)24h20，即当ab且h0时，所研究的函数为常数 ，12(ab)24h20时方程（*）,12

x1,2x3,4

，y1,2 (1h2(1h2(1h(221(2h2(2(2h(222f(xy)ax2by22hx

hy

by)

1

ax1hy11x1,hx1by11y1f(x,y)x2y2 1 1

f(x2,y2)1f(x3y3f(x4y42f点取四个(xiyi)(i1,2,3,4f(x1,y1)f(x2,y2)1，f(x3,y3)f(x4,y4)2x

,y

ax2by22hxy取最小值1xx3,4,yy3,4f(x,y取最大值21Lagrange1L(x,y,z)x2y2z2[(x2y2z2)2a2x2b2y2c2z212(lxmy令

2x2[2(x2y2z2)xa2x]l01212

2y2[2(x2y2z2)yb2y]

01L122 2z2[2(x2y2z2)zc2z]1L122

0(x2y2z2)2a2x2b2y2c2z2lxmynz0由条件(x2y2z22a2x2b2y2c2z2，得到a2x2b2y2c2a2x2b2y2c2zfx2y2z2ga2x2b2y2c2z2LagrangeL(xyz)a2x2b2y2c2z2(lxmynz

2a2xl02b2ym02c2zn0lxmynz0得唯一稳定点(0,0,0，很显然在点(0,0,0f(0,0,00fxmynzpxyzaa0m0n0p0x0y0z0解LagrangeL(xyz)xmynzp(xyza

mxm1ynzp0nxmyn1zp0pxmynzp10xyza

ynz

m

n

zxy

zkpaxyzk(mnp)k

mn

x mn

mn

mn

P0fxyzaxyaz0

xzay0

yzaxPf0f

mn

，y

mn

，z

mnmmnnppamnff(P0)(mnp)mnpz1(xnynxyl（l0n1）2a0b0n1(a2

an．2解L(xy)1(xnyn(xyl2

2020

xyl2xylfx0y0l,l2较

与边界点(0l),(l,0)f(0,l)

f(l,0)1ln2

(l)22

f(l,

l)（n12

fz(x,y)1(xnyn2

，当xy

n(n2

xn2

(l2

（xylx0y0时下面证明

a2

an2

（a0b0n1时ab0a0b0ab不同时为0abl,，则l0

an2

(l2

(ab)n2解xyz，则体积Vxyzs2(xyxzyz)s（常数L(xyz)xyz[2(xyxzyzsLxyz2(yz)0L xz2(xz)0LL xy2(xy)0L2(xyxzyz)ss6xyz s6解xyzs2(xyxz其体积为V（常数xyzVLagrange

L(x,y,z)2(xyxzyz)(xyzV)Lx2(yz)yz0L 2(xz)xz0LL 2(xy)xy0L3xyz3解r（常数SSr(xyz)，满足条件arctanx

arctany

arctanz

2

LagrangeL(xyz)r(xyzx

) Lr r r

x2ry2rz2r

000arctan

rarctanr

,xyzxyz

3r解xyxyax2 yLagrange

A

L(x,y)22

y

(xya)，L x0L L L

0x

4

，y

．4

xy4，且其中长是一段围成正方形，短的一段围成圆时，所围正方解设二平面交线上的点为(xyzx2yx2y2zd2dda1xb1yc1zd10，a2xb2yc2zd20下的最小值点与2在相Lagrange函数x2y2zL(x,y,z) 1(a1xb1yc1zd1)2(a2xb2yc2zd2)2令Lxxa11a220Lybb0 1 2

zc11c220axbycz

0 xyz

12xyza2b2c2Aa2b2c2

，a

b

c

B

1 1 1x

B2

x0y(b1d2b2d1)Bb1d1A2b2d2A1x0AAB 1z(c1d2c2d1)Bc1d1A2c2d2A1zAAB 1x2yx2y2z yx2xy1解设抛物线上的点为(xy，直线上的点为(xyyx2，x

(xx)(xx)2(yy xx2y2 L

y1x2

y

x2

1)Lx1L

x1

11011(xx)(xx)2(yy)2 1