题目
164 设f(x,y)有二阶连续偏导数, (x,y)=f((e)^xy,(x)^2+(y)^2), 且 f(x,y)=1-x-y+-|||-(sqrt ({(x-1))^2+(y)^2}), 证明g(x,y)在点(0,0)处取得极值,判断此极值是极大值还是极小值,并-|||-求出此极值.
题目解答
答案
解析
步骤 1：确定g(x,y)在点(0,0)处的偏导数
由题设，$g(x,y)=f({e}^{xy},{x}^{2}+{y}^{2})$，且$f(x,y)=1-x-y+o(\sqrt{(x-1)^2+y^2})$。首先，我们需要计算$g(x,y)$在点(0,0)处的一阶偏导数$g_x(0,0)$和$g_y(0,0)$。
步骤 2：计算$g_x(0,0)$和$g_y(0,0)$
根据链式法则，$g_x(x,y)=f_1'(e^{xy},x^2+y^2)\cdot ye^{xy}+f_2'(e^{xy},x^2+y^2)\cdot 2x$，$g_y(x,y)=f_1'(e^{xy},x^2+y^2)\cdot xe^{xy}+f_2'(e^{xy},x^2+y^2)\cdot 2y$。将$x=0$和$y=0$代入，得到$g_x(0,0)=0$和$g_y(0,0)=0$。
步骤 3：计算$g(x,y)$在点(0,0)处的二阶偏导数
根据链式法则，$g_{xx}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot y^2e^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot 2xye^{xy}+f_{21}''(e^{xy},x^2+y^2)\cdot 2ye^{xy}+f_{22}''(e^{xy},x^2+y^2)\cdot 4x^2$，$g_{xy}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot xye^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot (e^{xy}+2xye^{xy})+f_{21}''(e^{xy},x^2+y^2)\cdot (e^{xy}+2xye^{xy})+f_{22}''(e^{xy},x^2+y^2)\cdot 4xy$，$g_{yy}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot x^2e^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot 2xye^{xy}+f_{21}''(e^{xy},x^2+y^2)\cdot 2xe^{xy}+f_{22}''(e^{xy},x^2+y^2)\cdot 4y^2$。将$x=0$和$y=0$代入，得到$g_{xx}(0,0)=2f_2'(1,0)=-2$，$g_{xy}(0,0)=f_1'(1,0)=-1$，$g_{yy}(0,0)=2f_2'(1,0)=-2$。
步骤 4：判断极值
根据二阶偏导数的符号，$g_{xx}(0,0)=-2<0$，$g_{yy}(0,0)=-2<0$，$g_{xy}(0,0)=-1$，$g_{xx}(0,0)g_{yy}(0,0)-g_{xy}^2(0,0)=3>0$，因此$g(x,y)$在点(0,0)处取得极大值。
步骤 5：求出极值
由于$g(0,0)=f(1,0)=0$，因此$g(x,y)$在点(0,0)处的极大值为0。
由题设，$g(x,y)=f({e}^{xy},{x}^{2}+{y}^{2})$，且$f(x,y)=1-x-y+o(\sqrt{(x-1)^2+y^2})$。首先，我们需要计算$g(x,y)$在点(0,0)处的一阶偏导数$g_x(0,0)$和$g_y(0,0)$。
步骤 2：计算$g_x(0,0)$和$g_y(0,0)$
根据链式法则，$g_x(x,y)=f_1'(e^{xy},x^2+y^2)\cdot ye^{xy}+f_2'(e^{xy},x^2+y^2)\cdot 2x$，$g_y(x,y)=f_1'(e^{xy},x^2+y^2)\cdot xe^{xy}+f_2'(e^{xy},x^2+y^2)\cdot 2y$。将$x=0$和$y=0$代入，得到$g_x(0,0)=0$和$g_y(0,0)=0$。
步骤 3：计算$g(x,y)$在点(0,0)处的二阶偏导数
根据链式法则，$g_{xx}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot y^2e^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot 2xye^{xy}+f_{21}''(e^{xy},x^2+y^2)\cdot 2ye^{xy}+f_{22}''(e^{xy},x^2+y^2)\cdot 4x^2$，$g_{xy}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot xye^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot (e^{xy}+2xye^{xy})+f_{21}''(e^{xy},x^2+y^2)\cdot (e^{xy}+2xye^{xy})+f_{22}''(e^{xy},x^2+y^2)\cdot 4xy$，$g_{yy}(x,y)=f_{11}''(e^{xy},x^2+y^2)\cdot x^2e^{2xy}+f_{12}''(e^{xy},x^2+y^2)\cdot 2xye^{xy}+f_{21}''(e^{xy},x^2+y^2)\cdot 2xe^{xy}+f_{22}''(e^{xy},x^2+y^2)\cdot 4y^2$。将$x=0$和$y=0$代入，得到$g_{xx}(0,0)=2f_2'(1,0)=-2$，$g_{xy}(0,0)=f_1'(1,0)=-1$，$g_{yy}(0,0)=2f_2'(1,0)=-2$。
步骤 4：判断极值
根据二阶偏导数的符号，$g_{xx}(0,0)=-2<0$，$g_{yy}(0,0)=-2<0$，$g_{xy}(0,0)=-1$，$g_{xx}(0,0)g_{yy}(0,0)-g_{xy}^2(0,0)=3>0$，因此$g(x,y)$在点(0,0)处取得极大值。
步骤 5：求出极值
由于$g(0,0)=f(1,0)=0$，因此$g(x,y)$在点(0,0)处的极大值为0。