# 2018年考研数二第19题解析：条件极值、拉格朗日乘数法

## 解析

• 这三个图形的面积之和为：

$$s = \pi x^{2} + y^{2} + \frac{1}{2} \cdot z \cdot \sqrt{z^{2} – (\frac{z}{2})^{2}} \Rightarrow$$

$${\color{Red} s = \pi x^{2} + y^{2} + \frac{\sqrt{3}}{4}z^{2}}.(x > 0, y > 0, z > 0)$$

• 铁丝的长度为：

$${\color{Red} l = 2 = 2 \pi x + 4y + 3z}.(x > 0, y > 0, z > 0)$$

$$f(x,y,z) = \pi x^{2} + y^{2} + \frac{\sqrt{3}}{4}z^{2}.(x > 0, y > 0, z > 0)$$

$${\color{Red} L(x,y,z,\lambda) = \pi x^{2} + y^{2} + \frac{\sqrt{3}}{4}z^{2} + \lambda(2 \pi x + 4y + 3z – 2)}.$$

$${\color{Red} \left\{\begin{matrix} \frac{\partial L}{\partial x} = 2 \pi x + 2 \pi \lambda = 0;\\ \frac{\partial L}{\partial y} = 2y + 4 \lambda = 0;\\ \frac{\partial L}{\partial z} = \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ \frac{\partial L}{\partial \lambda} = 2 \pi x + 4y + 3z – 2 = 0 \end{matrix}\right.} \Rightarrow$$

$$\left\{\begin{matrix} x + \lambda = 0;\\ 2y + 4 \lambda = 0;\\ \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ 2 \pi x + 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} {\color{Red} \lambda = -x};\\ 2y + 4 \lambda = 0;\\ \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ 2 \pi x + 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} 2y – 4 x = 0;\\ \frac{\sqrt{3}}{2} z – 3 x = 0;\\ 2 \pi x + 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} {\color{Red} y = 2x};\\ \frac{\sqrt{3}}{2} z – 3 x = 0;\\ 2 \pi x + 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} \frac{\sqrt{3}}{2} z – 3 x = 0;\\ 2 \pi x + 8x + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} {\color{Red} x = \frac{\sqrt{3}}{6} z};\\ (2 \pi + 8) x + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$(2 \pi + 8) \frac{\sqrt{3}}{6} z + 3z – 2 = 0 \Rightarrow$$

$$[(2 \pi + 8) \frac{\sqrt{3}}{6} + 3] z = 2 \Rightarrow$$

$$[(2 \pi + 8) \frac{\sqrt{3}}{3} + 3] z = 2 \Rightarrow$$

$${\color{Red} z_{0} = \frac{2 \sqrt{3}}{\pi + 4 + 3 \sqrt{3}}}.$$

$$x_{0} = \frac{\sqrt{3}}{6} \cdot z_{0} = \frac{1}{\pi + 4 + 3 \sqrt{3}}.$$

$$y_{0} = 2 \cdot x_{0} = \frac{2}{\pi + 4 + 3 \sqrt{3}}.$$

$${\color{Red} \left\{\begin{matrix} x_{0} = \frac{1}{\pi + 4 + 3 \sqrt{3}};\\ y_{0} = \frac{2}{\pi + 4 + 3 \sqrt{3}};\\ z_{0} = \frac{2 \sqrt{3}}{\pi + 4 + 3 \sqrt{3}}. \end{matrix}\right.}$$

$$L_{1}(x,y,z,\lambda) = y^{2} + \frac{\sqrt{3}}{4}z^{2} + \lambda(4y + 3z – 2).$$

$$\left\{\begin{matrix} \frac{\partial L_{1}}{\partial y} = 2y + 4 \lambda = 0;\\ \frac{\partial L_{1}}{\partial z} = \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ \frac{\partial L_{1}}{\partial \lambda} = 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} 2y + 4 \lambda = 0;\\ \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} y = – 2 \lambda;\\ \frac{\sqrt{3}}{2} z + 3 \lambda = 0;\\ 4y + 3z – 2 = 0 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} x_{1} = 0;\\ y_{1} = \frac{2}{4 + 3 \sqrt{3}};\\ z_{1} = \frac{2 \sqrt{3}}{4 + 3 \sqrt{3}} \end{matrix}\right. \Rightarrow$$

$$f(x_{1}, y_{1}, z_{1}) = \frac{1}{4 + 3 \sqrt{3}}.$$

$$f(x_{0}, y_{0}, z_{0}) = \frac{1}{\pi + 4 + 3 \sqrt{3}}.$$

$$\frac{1}{\pi + 4 + 3 \sqrt{3}} < \frac{1}{4 + 3 \sqrt{3}}.$$

$$f(x_{0}, y_{0}, z_{0}) = \frac{1}{\pi + 4 + 3 \sqrt{3}}.$$