7.6 多元函数的极值2

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1、用拉格朗日乘数法求zf(x,y)在条件(x,y)0下的极值,其步骤为:(1):构造拉格朗日函数L(x,y,)f(x,y)(x,y),Lx(x,y,)fx(x,y)x(x,y)0(2):由Ly(x,y,)fy(x,y)y(x,y)0求出L(x,y,)的驻点(x0,y0,0).L(x,y,)(x,y)02(3):若dL0,则(x,y,)为L(x,y,)的极小值点,000(x0,y0,0)故(x,y)为zf(x,y)在条件(x,y)0下的极小值点;002若dL0,则(x,y,)为L

2、(x,y,)的极大值点,000(x0,y0,0)故(x,y)为zf(x,y)在条件(x,y)0下的极大值点;002若dL0,则需要进一步判定.(x0,y0,0)2222dLL(x,y,)dxL(x,y,)dyL(x,y,)dxxyy2L(x,y,)dxdy2L(x,y,)dxd2L(x,y,)dyd.xyxyL(x,y,)f(x,y)(x,y)L(x,y,)(x,y),xxxxxL(x,y,)f(x,y)(x,y)L(x,y,)(x,

3、y),yyyyyL(x,y,)(x,y)L(x,y,)0;222dLL(x,y,)dxL(x,y,)dyxx000yy000(x0,y0,0)2L(x,y,)dxdy2(x,y)dxd2(x,y)dyd.xy000x00y00(x,y)x00(a):若(x,y)0,则dydxy00(x,y)y002(x,y)dxd2(x,y)dydx00y00(x,y)x002(x,y)dxd2(x,y)d[dx]0x00y00(x,y)y00222dL

4、L(x,y,)dxL(x,y,)dy2L(x,y,)dxdyxx000yy000xy000(x0,y0,0)2x(x0,y0)2L(x,y,)dxL(x,y,)[dx]xx000yy000(x,y)y00(x,y)x002L(x,y,)dx[dx]xy000(x,y)y00x(x0,y0)2Lxx(x0,y0,0)Lyy(x0,y0,0)[]y(x0,y0)x(x0,y0)22Lxy(x0,y0,0)[]dxy(x0,y0)22dL为d

5、x的二次齐次式.其中dx0.(x0,y0,0)(x,y)y00(b):若(x,y)0,则dxdyx00(x,y)x002(x,y)dxd2(x,y)dydx00y00(x,y)y002(x,y)d[dy]2(x,y)dyd0.x00y00(x,y)x00222dLL(x,y,)dxL(x,y,)dy2L(x,y,)dxdyxx000yy000xy000(x0,y0,0)(x,y)2y0022dLL(x,y,)[dy]L(x,y,)dyxx000yy000

6、(x0,y0,0)(x,y)x00(x,y)y002L(x,y,)dy[dy]xy000(x,y)x00y(x0,y0)2y(x0,y0)2Lxx(x0,y0,0)[]Lyy(x0,y0,0)2Lxy(x0,y0,0)[]dyx(x0,y0)x(x0,y0)例:求zf(x,y)xy1在条件(x1)(y1)1下的极值,其中x0,y0.解:由(x1)(y1)1xyxy0.构造拉格朗日函数L(x,y,)xy1(xyxy),Lx(x,y,)yy

7、0(1)yxy则Ly(x,y,)xx0(1)xxyxy0L(x,y,)xyxy0xyxy0x0(舍去),x2(x,y,)(2,2,2)为L(x,y,)的驻点.2由于L0,L1,L0,故dL2(12)dxdy2dxdy.xxxyyy(2,2,2)由于(x,y)xyxy(x,y)y1,(x,y)x1.xy22122dL2dxdy2[]dx2dx0.(2,2,2)