﻿ 請教一題橢圓問題(頁 1) - 高中的數學 - IV：線性代數 - Math Pro 數學補給站

## Math Pro 數學補給站's Archiver

### 請教一題橢圓問題

x^2-xy+4y^2=1  , 求x^2+4y^2的最大與最小值

x^2-xy+4y^2=1  , 求x^2+4y^2的最大與最小值

$$x^2-xy+4y^2=1$$ 的 $$\delta = (-1)^2 - 4\cdot 1\cdot 4 <0$$，圖形為橢圓或其退化的情形.

$\frac{{\partial f}}{{\partial x}} = \frac{{\partial f}}{{\partial y}} = \frac{{\partial f}}{{\partial \lambda }} = 0,$

$2x + \lambda\left(2x -y\right) =0 .......（1）$
$8y + \lambda\left(-x+8y\right)=0 .......（2）$
$x^2 -xy +4y^2 - 1=0 .......（3）$

$⇒ 2\left(1+\lambda\right) x= \lambda y 且 8\left(1+\lambda\right)y = \lambda x .......（＊）$

$16\left(1+\lambda\right)^2xy = \lambda^2 xy$
$⇒ xy\left(16\left(1+\lambda\right)^2 - \lambda^2\right)=0$
$⇒ xy\left(3\lambda+4\right)\left(5\lambda+4\right)=0$
$⇒ x=0 或 y=0 或 \lambda = -\frac{4}{5} 或 \lambda = -\frac{4}{3}$

1. 若 $$\lambda = -\frac{4}{5}$$ ，則可以解得 $$x^2 = \frac{4}{10},\, y^2 = \frac{1}{10} ⇒ x^2 + 4y^2 = \frac{4}{5}.$$

2. 若 $$\lambda = -\frac{4}{3}$$ ，則可以解得 $$x^2 = \frac{2}{3},\, y^2 = \frac{1}{6} ⇒ x^2 + 4y^2 = \frac{4}{3}.$$

$x^2-xy+4y^2=1 ⇒ x^2 + 4y^2 = 1+xy ........（※）$

$\frac{x^2 + 4y^2}{2}\geq \sqrt{x^2 \cdot 4y^2}$
$⇒ \frac{1+xy}{2}\geq 2 \left|xy\right|$
$⇒ -\frac{1+xy}{2}\leq 2 xy \leq \frac{1+xy}{2}$
$⇒ -\frac{1}{5}\leq xy \leq \frac{1}{3}$
$⇒ \frac{4}{5}\leq 1+xy \leq \frac{4}{3}$

$⇒ \frac{4}{5}\leq x^2+4y^2 \leq \frac{4}{3}$ 請問式中
-1/5<=xy<=1/3
-1/5 與1/3怎麼得到的