习题1.9

第 1 题

研究下列函数的极值.

(1) \(z = x^3 + y^3 - 3x^2 - 3y^2\);

(2) \(z = e^{2x}(x + y^2 + 2y)\);

(4) \(z = x_1 + \dfrac{x_2}{x_1} + \dfrac{x_3}{x_2} + \cdots + \dfrac{x_n}{x_{n - 1}} + \dfrac{2}{x_n}, \quad x_i > 1\);

(5) \(u = x + \dfrac{y^2}{4x} + \dfrac{z^2}{y} + \dfrac{2}{z}, \quad x, y, z > 0\).

解

(1) \(\dfrac{\partial z}{\partial x} = 3x^2 - 6x, \dfrac{\partial z}{\partial y} = 3y^2 - 6y\). Hesse 矩阵为 \(H = \begin{pmatrix}6x - 6 & 0\\ 0 & 6y - 6\end{pmatrix}\). 驻点有 \((0, 0), (0, 2), (2, 0), (2, 2)\). 代入 Hesse 矩阵判断正定性、负定性知极小值点为 \((2, 2)\), 极大值点为 \((0, 0)\). 因此极小值为 \(z(2, 2) = -8\), 极大值为 \(z(0, 0) = 0\).

(2) \(\dfrac{\partial z}{\partial x} = e^{2x}(2x + 2y^2 + 4y + 1), \dfrac{\partial z}{\partial y} = e^{2x}(2y + 2)\). Hesse 矩阵为 \(H = \begin{pmatrix}e^{2x}(4x + 4y^2 + 8y + 4) & (4y+4)e^{2x}\\ (4y+4)e^{2x} & 2e^{2x}\end{pmatrix}\). 驻点为 \((\dfrac{1}{2}, -1)\). 代入 Hesse 矩阵, 为正定的. 因此有极小值为 \(z\left(\dfrac{1}{2}, -1\right) = -\dfrac{e}{2}\).

(4) \(\dfrac{\partial z}{\partial x_1} = 1 - \dfrac{x_2}{x_1}\), \(\dfrac{\partial z}{\partial x_i} = \dfrac{1}{x_{i - 1}} - \dfrac{x_{i + 1}}{x_i^2}, \quad (2 \le i \le n - 1)\), \(\dfrac{\partial z}{\partial x_n} = \dfrac{1}{x_{n - 1}} - \dfrac{2}{x_n^2}\). 因此 Hesse 矩阵为

\[ \begin{pmatrix} \dfrac{2x_2}{x_1^3} & -\dfrac{1}{x_1^2} & 0 & 0 & \cdots & 0 \\ -\dfrac{1}{x_1^2} & \dfrac{2x_3}{x_2^3} & -\dfrac{1}{x_2^2} & 0 & \cdots & 0 \\ 0 & -\dfrac{1}{x_2^2} & \dfrac{2x_4}{x_3^3} & -\dfrac{1}{x_3^2} & \cdots & 0 \\ 0 & 0 & -\dfrac{1}{x_3^2} & \dfrac{2x_5}{x_4^3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \dfrac{4}{x_n^3} \end{pmatrix} \]

驻点为 \((2^{\frac{1}{n+1}}, 2^{\frac{2}{n+1}}, \cdots, 2^{\frac{n}{n+1}})\). 代入 Hesse 矩阵知其为正定的, 因此有极小值 \(z(2^{\frac{1}{n+1}}, 2^{\frac{2}{n+1}}, \cdots, 2^{\frac{n}{n+1}}) = (n+1)2^{\frac{1}{n+1}}\).

(5) \(\dfrac{\partial u}{\partial x} = 1 - \dfrac{y^2}{4x^2}, \dfrac{\partial u}{\partial y} = \dfrac{y}{2x} - \dfrac{z^2}{y^2}, \dfrac{\partial u}{\partial z} = \dfrac{2z}{y} - \dfrac{2}{z^2}\). Hesse 矩阵为 \(\begin{pmatrix}\dfrac{y^2}{2x^3} & -\dfrac{y}{2x^2} & 0\\ -\dfrac{y}{2x^2} & \dfrac{1}{2x} + \dfrac{2z^2}{y^3} & -\dfrac{2z}{y^2} \\ 0 & -\dfrac{2z}{y^2} & \dfrac{2}{y} + \dfrac{4}{z^3}\end{pmatrix}\). 驻点为 \(\left(\dfrac{1}{2}, 1, 1\right)\) 和 \(\left(-\dfrac{1}{2}, -1, -1\right)\). 代入 Hesse 矩阵知极小值点为 \((\dfrac{1}{2}, 1, 1)\). 因此极小值为 \(u(\dfrac{1}{2}, 1, 1) = 4\).

第 2 题

函数 \(z = z(x, y)\) 由 \(2x^2 + 2y^2 + z^2 + 8xz - z + 8 = 0\) 确定, 求 \(z = z(x, y)\) 的极值.

解

记 \(F(x, y, z) = 2x^2 + 2y^2 + z^2 + 8xz - z + 8\). 因此 \(\dfrac{\partial z}{\partial x} = -\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial z}} = -\dfrac{4x + 8z}{2z + 8x - 1}, \dfrac{\partial z}{\partial y} = -\dfrac{\dfrac{\partial F}{\partial y}}{\dfrac{\partial F}{\partial z}} = \dfrac{4y}{2z + 8x - 1}\). 因此在极值点有 \(\begin{cases}4x_0 + 8z_0 = 0, \\ y_0 = 0.\end{cases}\) 代入方程有 \(2(-2z_0)^2 + z_0^2 + 8(-2z_0)z_0 - z_0 + 8 = 0\), 化简得 \(7z_0^2 + z_0 - 8 = 0\). 解得 \(z_{01} = 1. z_{02} = -\dfrac{8}{7}\). 经检验 Hesse 矩阵也满足条件, 所以极小值为 \(1\), 极大值为 \(-\dfrac{8}{7}\).

第 4 题

求下列函数在给定区域的最值.

(2) \(z = xy(4 - x - y), (x, y) \in \{x + y \le 6, y \ge 0, x \ge 1\}\).

解

由于 \(\dfrac{\partial z}{\partial x} = 4y - y^2 - 2xy, \dfrac{\partial z}{\partial y} = 4x - x^2 - 2xy\), 所以驻点满足方程

\[ \begin{cases} 4y - y^2 - 2xy = 0 \\ 4x - x^2 - 2xy = 0 \end{cases} \]

解得 \((x, y) = (0, 0), (0, 4), \left(\dfrac{4}{3}, \dfrac{4}{3}\right), (4, 0)\). 再加上 \(x \ge 1\) 的条件后, 只剩下 \(z\left(\dfrac{4}{3}, \dfrac{4}{3}\right) = \dfrac{64}{27}\) 和 \(z(4, 0) = 0\).

再考虑边界上的极值. 当 \(y = 0\) 时 \(z \equiv 0\). 当 \(x = 1\) 时 \(z = y(3 - y) \in \left[0, \dfrac{9}{4}\right]\). 当 \(x + y = 6\) 时 \(z = -2xy \in [-18, 0]\).

综上可得 \(z\) 有最大值 \(\dfrac{64}{27}\), 最小值 \(-18\).

第 5 题

证明下列各题

(1) 设 \(D\) 为 \(\mathbb{R}^2\) 中的有界闭区域. \(f(x, y)\) 在 \(D\) 上连续, 在 \(D\) 内可微, 且满足方程 \(\dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} = kf(x, y)(k > 0)\), 若在 \(D\) 的边界上有 \(f(x, y) = 0\), 试证 \(f(x, y)\) 在 \(D\) 上恒为 \(0\).

(2) 设 \(u(x, y)\) 在 \(x^2 + y^2 \le 1\) 上连续, 在 \(x^2 + y^2 < 1\) 满足 \(\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2u}{\partial y^2} = u\). 且在 \(x^2 + y^2 = 1\) 上, \(u(x, y) \ge 0\). 证明: 当 \(x^2 + y^2 \le 1\) 时, \(u(x, y) \ge 0\).

证明

(1) 反证法. 假设 \(\exists P(x_0, y_0)\) 使得 \(f(x_0, y_0) \neq 0\). 不妨设 \(f(x_0, y_0) > 0\). 则由题知 \(\dfrac{\partial f}{\partial x}\bigg\vert_P + \dfrac{\partial f}{\partial y}\bigg\vert_P = kf(x_0, y_0) > 0\). 因此 \(\dfrac{\partial f}{\partial x}\bigg\vert_P\) 和 \(\dfrac{\partial f}{\partial y}\bigg\vert_P\) 中至少有一个为正数. 不妨设 \(\dfrac{\partial f}{\partial x}\bigg\vert_P> 0\). 则 \(f(x, y)\) 从 \(f(x_0, y_0)\) 沿着 \(x\) 方向单调递增. 而 \(f(x_0, y_0) > 0\), 所以沿着该方向到边界时, 有边界点函数值也为正数, 矛盾. 因此 \(f(x, y) \equiv 0\).

(2) 反证法. 假设 \(\exists P(x_0, y_0)\) 使得 \(u(x_0, y_0) < 0\), 且 \(u\) 在 \(P(x_0, y_0)\) 处取得极小值. 则在极小值点 \(P(x_0, y_0)\) 处的 Hesse 矩阵的迹为 \(\dfrac{\partial^2 u}{\partial x^2}\bigg\vert_P + \dfrac{\partial^2u}{\partial y^2}\bigg\vert_P > 0\). 这与题目 \(\dfrac{\partial^2 u}{\partial x^2}\bigg\vert_P + \dfrac{\partial^2u}{\partial y^2}\bigg\vert_P = u(x_0, y_0) < 0\) 矛盾. 因此当 \(x^2 + y^2 \le 1\) 时, \(u(x, y) \ge 0\).

第 6 题

证明函数 \(z(x, y) = (1 + e^y)\cos x - ye^y\) 有无穷多个极大值点而无极小值.

证明

由于 \(\dfrac{\partial z}{\partial x} = -(1 + e^y)\sin x, \dfrac{\partial z}{\partial y} = (\cos x - 1 - y)e^y\), 所以 Hesse 矩阵为

\[ H = \begin{pmatrix}\dfrac{\partial^2 z}{\partial x^2} & \dfrac{\partial ^2z}{\partial x\partial y} \\ \dfrac{\partial^2z}{\partial x\partial y} & \dfrac{\partial^2z}{\partial y^2}\end{pmatrix} = \begin{pmatrix}-(1+e^y)\cos x & -e^y\sin x \\ -e^y\sin x & (\cos x - y - 2)e^y\end{pmatrix} \]

且驻点为 \((2k\pi, 0)\) 和 \(((2k+1)\pi, 2)\), 其中 \(k \in \mathbb{Z}\). 将 \((2k\pi, 0)\) 代入 \(H\), 有 \(H_1 = \begin{pmatrix}-2 & 0 \\ 0 & -1\end{pmatrix}\) 负定. 因此 \((2k\pi, 0), k \in \mathbb{Z}\) 为极大值点, 且有无穷多个. 将 \(((2k + 1)\pi, 2)\) 代入 \(H\), 有 \(H_2 = \begin{pmatrix}e^2 + 1 & 0 \\ 0 & -5e^2\end{pmatrix}\) 不定. 因此 \(((2k + 1)\pi, 2), k \in \mathbb{Z}\) 不为极值点. 所以没有极小值点.

第 7 题

求下列函数在给定条件下的条件极值.

(1) \(\begin{cases}z = x^2 + y^2, \\ \text{s.t.} \dfrac{x}{a} + \dfrac{y}{b} = 1;\end{cases}\)

(2) \(\begin{cases}u = x - 2y + 2z, \\ \text{s.t.} x^2 + y^2 + z^2 = 1;\end{cases}\)

解

(1) \(L = x^2 + y^2 + \lambda\left(\dfrac{x}{a} + \dfrac{y}{b} - 1\right)\). \(L_x' = 2x + \dfrac{\lambda}{a}, L_y' = 2y + \dfrac{\lambda}{b}\). 因此有方程组 \(\begin{cases}2x + \dfrac{\lambda}{a} = 0 \\2y + \dfrac{\lambda}{b} = 0 \\\dfrac{x}{a} + \dfrac{y}{b} = 1\end{cases}\), 解得 \(\begin{cases}x = \dfrac{ab^2}{a^2 + b^2} \\y = \dfrac{a^2b}{a^2 + b^2} \\\lambda = -\dfrac{2a^2b}{a^2 + b^2}\end{cases}\). 此时有 \(z = \dfrac{a^2b^2}{a^2 + b^2}\) 为极小值.

(2) \(L = x - 2y + 2z + \lambda(x^2 + y^2 + z^2 - 1)\). \(L_x' = 1 + 2\lambda x, L_y' = -2 + 2\lambda y, L_z' = 2 + 2\lambda z\). 因此有方程组 \(\begin{cases}1 + 2\lambda x = 0 \\ -2 + 2\lambda y = 0 \\ 2 + 2\lambda z = 0 \\ x^2 + y^2 + z^2 = 1\end{cases}\). 解得 \(\begin{cases}x = -\dfrac{1}{3} \\ y = \dfrac{2}{3} \\ z = -\dfrac{2}{3} \\ \lambda = \dfrac{3}{2}\end{cases}\) 或 \(\begin{cases}x = \dfrac{1}{3} \\ y = -\dfrac{2}{3} \\ z = \dfrac{2}{3} \\ \lambda = -\dfrac{3}{2}\end{cases}\). 因此有极小值 \(u\left(-\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{2}{3}\right) = -3\), 极大值 \(u\left(\dfrac{1}{3}, -\dfrac{2}{3}, \dfrac{2}{3}\right) = 3\).

第 8 题

求函数 \(u = x^2 + 2y^2 + z^2 - 2xy - 2yz\) 在区域 \(x^2 + y^2 + z^2 \le 4\) 内的最值.

解

\[\begin{enumerate} \item 考虑内点处的极值. $\dfrac{\partial u}{\partial x} = 2x - 2y, \dfrac{\partial u}{\partial y} = 4y - 2x - 2z, \dfrac{\partial u}{\partial z} = 2z - 2y$. 因此驻点满足 $x = y = z$. 此时 $u = 0$. \item 考虑边界点处的极值. $L = x^2 + 2y^2 + z^2 - 2yz - 2xy + \lambda(x^2 + y^2 + z^2 - 4)$. $L_x' = 2x - 2y + 2\lambda x, L_y' = 4y - 2z - 2z + 2\lambda y, L_z' = 2z - 2y + 2\lambda z$. 因此有方程组 $\begin{cases}2x - 2y + 2\lambda x = 0 \\ 4y - 2z - 2z + 2\lambda y = 0 \\ 2z - 2y + 2\lambda z = 0 \\ x^2 + y^2 + z^2 = 4\end{cases}$. 解得 $(x, y, z) = (0, 0, 0), \left(\dfrac{2\sqrt{3}}{3}, \dfrac{2\sqrt{3}}{3}, \dfrac{2\sqrt{3}}{3}\right), \left(\pm\dfrac{\sqrt{6}}{3}, \mp\dfrac{2\sqrt{6}}{3}, \pm\dfrac{\sqrt{6}}{3}\right), (\pm\sqrt{2}, 0, \mp\sqrt{2}),$. 故有最大值 $u\left(\pm\dfrac{\sqrt{6}}{3}, \mp\dfrac{2\sqrt{6}}{3}, \pm\dfrac{\sqrt{6}}{3}\right) = 12$, 最小值 $u(k, k, k)(k \in \mathbb{R}) \equiv 0$. \end{enumerate}\]

第 9 题

求解下列问题.

(1) 求椭圆 \(x^2 + \dfrac{y^2}{4} = 1\) 上的点到直线 \(x + y = 4\) 的距离的最值;

(3) 求椭圆面 \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1\) 内接长方体的体积最值;

解

(1) 设椭圆上的某一点为 \(P(\cos \theta, 2\sin \theta), \theta \in (-\pi, \pi]\). 距离为 \(d = \dfrac{\left|\cos \theta + 2\sin \theta - 4\right|}{\sqrt{2}} = \dfrac{\left|4 - \sqrt{5}\cos\varphi\right|}{\sqrt{2}} \in [\dfrac{4\sqrt{2} - \sqrt{10}}{2}, \dfrac{4\sqrt{2} + \sqrt{10}}{2}]\). 因此最大值为 \(\dfrac{4\sqrt{2} + \sqrt{10}}{2}\), 最小值为 \(\dfrac{4\sqrt{2} - \sqrt{10}}{2}\).

(3) 易知该长方体关于原点中心对称. 设长方体的某一个顶点为 \(P(a\cos\theta\cos\varphi, b\cos\theta\sin\varphi, c\sin\theta)\). 则体积为 \(V = 8\left|xyz\right| = 8abc\cos^2\theta\sin\theta\sin\varphi\cos\varphi = 4abc\sin\theta(1 - \sin^2\theta)\sin2\varphi \in (0, \dfrac{2\sqrt{3}}{9}abc]\). 因此体积最大值为 \(\dfrac{8\sqrt{3}}{9}abc\), 无最小值.

第 10 题

求解下列问题.

(3) 已知矩形的周长为 \(2p\), 将它绕其一边旋转成的圆柱体的体积最大值为多少?

解

设矩形的一边长为 \(x\), 则邻边长为 \(p - x\). 绕长为 \(p - x\) 的边旋转而成的圆柱体体积为 \(\pi x^2 (p-x) = \dfrac{\pi}{2}x \cdot x \cdot (2p - 2x) \le \dfrac{\pi}{2}\left(\dfrac{x + x + 2p - 2x}{3}\right)^3 = \dfrac{4p^3\pi}{27}\). 因此圆柱体的体积最大值为 \(\dfrac{4p^3\pi}{27}\).

第 11 题

求解下列问题.

(1) 求函数 \(u = x^2y^2z^2\) 在约束条件 \(x^2 + y^2 + z^2 = a^2\) 下的最大值, 并证明:

\[ \sqrt[3]{x^2y^2z^2} \le \dfrac{x^2 + y^2 + z^2}{3}; \]

(2) 类似 (1), 证明: \(a_1, a_2, \cdots, a_n > 0\) 有

\[ \sqrt{\dfrac{a_1^2 + a_2^2 + \cdots + a_n^2}{n}} \ge \dfrac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n} \ge \dfrac{n}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_n}}. \]

证明

(1) 设 \(L = x^2y^2z^2 + \lambda(x^2 + y^2 + z^2 - a^2)\), 则

\[ \begin{aligned} L_x' & = 2xy^2z^2 + 2x\lambda = 0 \\ L_y' & = 2x^2yz^2 + 2y\lambda = 0 \\ L_z' & = 2x^2y^2z + 2z\lambda = 0 \end{aligned} \]

解得有极大值点满足 \(x^2 = y^2 = z^2 = \dfrac{a^2}{3}\), 因此有 \(x^2y^2z^2 \le \left(\dfrac{a^2}{3}\right)^3 = \left(\dfrac{x^2 + y^2 + z^2}{3}\right)^3\). 此即 \(\sqrt[3]{x^2y^2z^2} \le \dfrac{x^2 + y^2 + z^2}{3}\).

(2) 同理, 拓展至 \(n\) 元, 有 \(\sqrt{\dfrac{a_1^2 + a_2^2 + \cdots + a_n^2}{n}} \ge \dfrac{a_1 + a_2 + \cdots + a_n}{n}\). 而由 \(n\) 元均值不等式有 \(\dfrac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}\). 将 \(a_1 = \dfrac{1}{b_1}, a_2 = \dfrac{1}{b_2}, \cdots, a_n = \dfrac{1}{b_n}\) 代入有 \(\dfrac{\dfrac{1}{b_1} + \dfrac{1}{b_2} + \cdots + \dfrac{1}{b_n}}{n}\ge \dfrac{1}{\sqrt[n]{b_1b_2\cdots b_n}}\), 不等式两边同时取倒数, 有 \(\sqrt[n]{b_1b_2\cdots b_n} \ge \dfrac{n}{\dfrac{1}{b_1} + \dfrac{1}{b_2} + \cdots + \dfrac{1}{b_n}}\). 此即 \(\sqrt[n]{a_1a_2\cdots a_n} \ge \dfrac{n}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_n}}\). 命题得证.

第 13 题

弹簧在弹性限度内的伸长 \(x\) 与所受拉力 \(y\) 满足关系式 \(y = a + bx\), 试根据下列数据确定弹性系数 \(b\).

\[\begin{table}[H] \centering %\caption{} \vspace{1em} \begin{tabular}{|c|c|c|c|c|} \hline $x_i/\text{cm}$ & 2.6 & 3.0 & 3.5 & 4.3 \\ \hline $y_i/\text{kg}$ & 0 & 1 & 2 & 3 \\ \hline \end{tabular} %\label{} \end{table}\]

解

由题可列出方程组 \(A\mathbf{x}=\mathbf{b}\):

\[ \begin{pmatrix} 1 & 2.6 \\ 1 & 3.0 \\ 1 & 3.5 \\ 1 & 4.3 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \\ 3 \end{pmatrix} \]

则 \(A^{\top}A = \begin{pmatrix}1 & 1 & 1 & 1 \\ 2.6 & 3.0 & 3.5 & 4.3\end{pmatrix}\begin{pmatrix}1 & 2.6 \\1 & 3.0 \\1 & 3.5 \\1 & 4.3\end{pmatrix}=\begin{pmatrix}4 & 13.4 \\ 13.4 & 46.5\end{pmatrix}\), \(A^{\top}\mathbf{b} = \begin{pmatrix}1 & 1 & 1 & 1 \\ 2.6 & 3.0 & 3.5 & 4.3\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 2 \\ 3\end{pmatrix}=\begin{pmatrix}6 \\ 22.9\end{pmatrix}\)

因此方程即

\[ \begin{pmatrix} 4 & 13.4 \\ 13.4 & 46.5 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 6 \\ 22.9 \end{pmatrix} \]

解得 \(a = -4.326, b = 1.739\).