方企勤 第五章 多元函数微分学 第6题
📝 题目
例 6 求由方程 $2{x}^{2} + {y}^{2} + {z}^{2} + {2xy} - {2x} - {2y} - {4z} + 4 = 0$ 所确定的函数 $z = z\left( {x,y}\right)$ 的极值.
💡 答案解析
解法 1 由隐函数求导,得
$$ \left\{ \begin{array}{l} {4x} + {2z}\frac{\partial z}{\partial x} + {2y} - 2 - 4\frac{\partial z}{\partial x} = 0, \\ {2y} + {2z}\frac{\partial z}{\partial y} + {2x} - 2 - 4\frac{\partial z}{\partial y} = 0. \end{array}\right. \tag{4.5} $$ (4.6)
令 $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0$ ,得方程组
$$ \left\{ \begin{array}{l} {2x} + y - 1 = 0 \\ y + x - 1 = 0 \end{array}\right. $$
由此求出临界点 $x = 0,y = 1$ . 再代入原方程,求出两个隐函数的值为
$$ {z}_{1} = {z}_{1}\left( {0,1}\right) = 1,\;{z}_{2} = {z}_{2}\left( {0,1}\right) = 3. $$
求二阶偏导数, 由 (4.5) 和 (4.6) 式, 得
$$ \left\{ \begin{array}{l} 4 + 2{\left( \frac{\partial z}{\partial x}\right) }^{2} + {2z}\frac{{\partial }^{2}z}{\partial {x}^{2}} - 4\frac{{\partial }^{2}z}{\partial {x}^{2}} = 0, \\ 2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y} + {2z}\frac{{\partial }^{2}z}{\partial x\partial y} + 2 - 4\frac{{\partial }^{2}z}{\partial x\partial y} = 0, \\ 2 + 2{\left( \frac{\partial z}{\partial y}\right) }^{2} + {2z}\frac{{\partial }^{2}z}{\partial {y}^{2}} - 4\frac{{\partial }^{2}z}{\partial {y}^{2}} = 0. \end{array}\right. \tag{4.7} $$ (4.8) (4.9)
用 $x = 0,y = 1,{z}_{1} = 1$ 代入上式,得
$$ {A}_{1} = \frac{{\partial }^{2}z}{\partial {x}^{2}} = 2 > 0,\;{B}_{1} = \frac{{\partial }^{2}z}{\partial x\partial y} = 1,\;{C}_{1} = \frac{{\partial }^{2}z}{\partial {y}^{2}} = 1, $$
${A}_{1}{C}_{1} - {B}_{1}^{2} = 1 > 0$ ,所以隐函数 ${z}_{1}\left( {x,y}\right)$ 在(0,1)点有极小值 1 . 用 $x =$ $0,y = 1,{z}_{2} = 3$ 代入 (4.7) 式 $\sim \left( {4.9}\right)$ 式,得 ${A}_{2} = - 2 < 0,{B}_{2} = - 1,{C}_{2}$ $= - 1,{A}_{2}{C}_{2} - {B}_{2}^{2} = 1 > 0$ . 所以隐函数 ${z}_{2}\left( {x,y}\right)$ 在(0,1)点有极大值 3 .
解法 2 取目标函数 $f\left( {x,y,z}\right) = z$ ,约束条件为原方程. 令
$$ \Phi \left( {x,y,z,\lambda }\right) = z + \lambda \left( {2{x}^{2} + {y}^{2} + {z}^{2} + {2xy} - {2x} - {2y} - {4z} + 4}\right) . $$
求导得
$$ \left\{ \begin{array}{l} {\Phi }_{x}^{\prime } = \lambda \left( {{4x} + {2y} - 2}\right) = 0, \\ {\Phi }_{y}^{\prime } = \lambda \left( {{2y} + {2x} - 2}\right) = 0, \\ {\Phi }_{z}^{\prime } = 1 + \lambda \left( {{2z} - 4}\right) = 0, \\ {\Phi }_{\lambda }^{\prime } = 2{x}^{2} + {y}^{2} + {z}^{2} + {2xy} - {2x} - {2y} - {4z} + 4 = 0. \end{array}\right. \tag{4.10} $$ (4.11) (4. 12) (4.13)
容易看出 $\lambda \neq 0$ ,所以由 (4.10) 和 (4.11) 式解出 $x = 0,y = 1$ . 再由 (4.13) 式解出 ${z}_{1} = 1,{z}_{2} = 3$ ,经 (4.12) 式解出
$$ {\lambda }_{1} = 1/2,\;{\lambda }_{2} = - 1/2. $$
$\Phi \left( {x,y,z,1/2}\right)$ 在(0,1,1)点的海色矩阵为
$$ \left\lbrack \begin{array}{lll} 2 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack $$
它的主对角线行列式
$$ {\Delta }_{1} = 2 > 0,\;{\Delta }_{2} = 1 > 0,\;{\Delta }_{3} = 1 > 0. $$
所以隐函数 ${z}_{1}\left( {x,y}\right)$ 在(0,1)点有极小值 1 . $\Phi \left( {x,y,z, - 1/2}\right)$ 在 (0,1,3)点的海色矩阵为
$$ \left\lbrack \begin{array}{rrr} - 2 & - 1 & 0 \\ - 1 & - 1 & 0 \\ 0 & 0 & - 1 \end{array}\right\rbrack $$
它的主对角线行列式 ${\Delta }_{1} = - 2 < 0,{\Delta }_{2} = 1 > 0,{\Delta }_{3} = - 1 < 0$ . 所以隐函数 ${z}_{2}\left( {x,y}\right)$ 在(0,1)点有极大值 3 .