讲高数17堂专题课多元函数极值与最值

定义zf(xyP(x0y0的某邻域内有定义，若对该邻域内任意的点P(x,y)均有f(xy)f(x0y0)（f(xyf(x0,y0则称(x0y0f(xy的极大值点（或极小值点f(x0y0f(xy的极大极值1（极值的必要条件）设zf(xy在点(x0y0存在偏导数，且(x0y0f(xyfx(x0,y0) fy(x0,y0)2(极值的充分条件）zf(xyP0x0y0的某邻域内有二阶连续偏导fx(x0y0fy(x0y00。记Afxx(x0,y0 Bfxy(x0,y0 Cfyy(x0,y0ACB20，则(x0y0f(xyA0，则(x0y0f(xyA0，则(x0y0f(xyACB20，则(x0y0f(xyACB20，则(x0y0f(xyf(xy)的zf(xyf(xyP1,LPk【注】二元函数zf(xy)（x2f(xx2zf(xy在条件(xy)0（1）构造日函数F(x,y,)f(x,y)(x, f(x, (x, x,yx,y及，则其中(x,y就是函数f(xy)在条件(x,y)0下的可能的极值点.nm个约束条件下的极值问题，如求uf(xyz在条件(x,y,z)0，(x,y,z)0下的极值，可构造日函F(x,y,z,,)f fx(x,y,z)(x,y,z)(x,y,z) f(x,y,z)(x,y,z)(x,y,z) fz(x,y,z)z(x,y,z)z(x,y,z)(x,y,z)xyz及，则其中(xyz（1)f(xyD（2)f(x,yD（3)【1zxy(3xy的极值点是（（A） （B） zxy(32xy)zz

x(32yx)(0,00,3),(3,0),(1,1都满足上式zxx2y,zyy2x,zxy32x2y.在(0,0)点 ACB290,无极值;在(0,3)点 ACB290,无极值;在(3,0) ACB290,无极值在(1,1)点 ACB230,有极值.【2f(xg(xf(0)0g(0)0fg(0)0zf(x)gy在点(0,0)处取得极小值的一个充分条件是（（A）f(0)0,g(0) （B）f(0)0,g(0)（C）f(0)0,g(0) （D）f(0)0,g(0)zf(x)g(y)，z

f(x)g(y)

f(x)g(y) . .2z f(x)g(

zf(x)g(

，在(0,0)处ACB2f(0)g(0)f(0)g(0)[ff(0)g(0)ff(0)0g(00f(00g(0)0f(0)g(0)f(0)g(0)0，Af(0)g(0)0.zf(x)gy在点(0,0)处取得极小值.故选（A）.3zf(xy的全微分dzayx2dxaxy2dya0f(x, (B)点(a,a)为极小值点(C)点(a,a)为极大值点 (D)是否有极值点与a的取值有关【解】由dzayx2dxaxy2dyzayx2,zax ayx2令axy2

xayACB23a20,A2a则点(a,a)为极大值点f(x,y)(x2f(x,y)(x2y2 x2

a(A)点(0,0)不是f(x,y)的极值 (B)点(0,0)是f(x,y)的极大点点(0,0)f(xy根据所给条件无法判断点(0,0)f(xy【解】由

ax2f(x,y)(x2y2 x2，fx2f(x,y)(x2y2 x2，

f(0,0)其中lim(x,y)x2x2 f(x,y) (x, x2x2x2 x2f(x,y)a()当a0f(xy在(0,0当a0f(xy在(0,0当a0f(xy在(0,0x2f(x,y)(2x (x2x2f(0,0)0,f(x,0)2xxx0f(x,00;x0f(x,0)0;则点(0,0f(xy的极值点.【5f(xy与(xy均为可微函数，且y(xy0已知(x0y0f(xy在约束条件(x,y)0下的一个极值点，下列选项正确的是（）.（A）若fx(x0,y0)0，则fy(x0,y0) （B）若fx(x0,y0)0，则fy(x0,y0)（Ｃ）若fx(x0y00fy(x0y00（Ｄ）fx(x0y00fy(x0y0【解】构造日函数F(x,y,)f(x,y)(x,y)，Fx(x,yFy(x,y)0得fx(x0,y0)x(x0,y0)fy(x0,y0)y(x0,y0)

fx(x0y00（1）式知0又y(xy0此时由（2）式可知fy(x0y00故应选6】f(xyx22y2ylny的极值 f(x,y)2x(2y2)，f(x,y)2x2ylny fx(x,y) 1 令f(xy0解得唯一驻点0,e yAf0,12(2y2 221xx e

1 2 eBf0,14xyxy

1 e

0,eCf0,1 2x 1

yy

0,1 eACB22e210A0.f0,1f(xy f0,11

e2

e e 【7f(xyfxy2(y1)ex,fx(x,0)(x1)ex,f(0,y)y22f(xyfx(x,0)x1)ex所以ex(x)x得(xxexfxy1)2ex对x积分 f(x,y)(y1)2ex(x1)ex(f(0,y)y22y所以y0f(x,y)(xy22f2y2)ex,fxy22y2)ex,f fx0,fy0,得驻点(0,1所ACB20,A0,f(0,1) 0 0,

C,xy 2u由题设xy0, y20,可知,B0,AC0,ACB2故函数u(xy)D内无极值点,因此,u(x,y)D的边界上取2】zf(xy的全微分dz2xdx2ydyf(1,1)2.f(xy D(xy

1上的最大值和最小值 【解1】由dz2xdx2ydy可 zf(x,y)x2y2再由f(1,1)2，得C2， zf(x,y)x2y2 x2x0,y2y0解得驻点(0,0)x2y24

1z

(4

z5x2 (1xx x其最大值为 3最小值为 2.再与f(0,0)2比较可知f(x,yx x 日乘数法求此函数在椭圆x2y2Lx2y22

1上的极值4 L

2y

yLx2y1 解得4个可能的极值点(0,2),(0,2),(1,0)和(1,0)又f(0,22,f(0,2)2,f(1,0)3,f(1,03，再与f(0,0)2比较，得f(xyD3，最小值为23】同解法一，得驻点(0,0).x2y24

cost,

2sin zf(x,y)x2y22cos2t4sin2t35sin2fmax3,fmin3zz(x,y的微分dz2x12y)dx12x4y)dyz(0,00，求zz(x,y)4x2y225上的最大值。由dz2x12y)dx12x4y)dyzx212xy2zx2x12yz令 12x4yzyF(x,y,)x212xy2y2(4x2y2Fx2x12y8xF4x2y225(14)x6y

6x(2)y

1则6

62

0,解得12,24 24

,4),P4

,4),z 比较得zmax4】求函数ux2y2z2zx2y2xyz4下的最大值与最小【解】作日函F(x,y,z,,)x2y2z2(x2y2z)(xyzFx2x2xF2y2y Fz2z Fx2 解方程组，得最小值为6.5】x3xyy31x0y0)x2x2L(x,y,)x2y2(x3xyy3 L2x(3x2y) L2y(3y2x) Lx3xyy31 3x2x0y0y3y2x即3xyyxxy)(xyyx或3xy(xyx0,y0yx代入③得2x3x210,即(2x2x1)(x1)0x22x22x2当x0,y1或x1,y0时x22 2【6】将长为2m的铁丝分成三段，以次围成圆、正方形与正三角形.三个图形的面积之S(x,y,z)x2y234 2x4y3z (x0,y0,z L(x,y,z,)x2y2

z2(2x4y3z343 L2x2 2y4 L 3z3 L2x4y3z2 2解 x0433,y041 S(x,y,z) .1 43

,z0 434343又当2x4y3z2