Question

10．設(shè)實(shí)數(shù)x，y滿(mǎn)足約束條件$\left\{\begin{array}{l}{x+y≥0}\\{x-y≤0}\\{x+3y≤3}\end{array}\right.$，則$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$的取值范圍是[$-\sqrt{2}$，-1）∪（-1，0]．

Answer 1

分析 由約束條件作出可行域，由$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$=$-\sqrt{\frac{（x-y）^{2}}{{x}^{2}+{y}^{2}}}=-\sqrt{\frac{{x}^{2}+{y}^{2}-2xy}{{x}^{2}+{y}^{2}}}$=$-\sqrt{1-\frac{2xy}{{x}^{2}+{y}^{2}}}=-\sqrt{1-\frac{2•\frac{y}{x}}{1+（\frac{y}{x}）^{2}}}$．令k=$\frac{y}{x}$，則$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$=$-\sqrt{1-\frac{2k}{1+{k}^{2}}}=-\sqrt{1-\frac{2}{k+\frac{1}{k}}}$．由圖求出k的范圍，再由基本不等式求得答案．

解答 解：由約束條件$\left\{\begin{array}{l}{x+y≥0}\\{x-y≤0}\\{x+3y≤3}\end{array}\right.$作出可行域如圖，

$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$=$-\sqrt{\frac{（x-y）^{2}}{{x}^{2}+{y}^{2}}}=-\sqrt{\frac{{x}^{2}+{y}^{2}-2xy}{{x}^{2}+{y}^{2}}}$=$-\sqrt{1-\frac{2xy}{{x}^{2}+{y}^{2}}}=-\sqrt{1-\frac{2•\frac{y}{x}}{1+（\frac{y}{x}）^{2}}}$．
令k=$\frac{y}{x}$，則$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$=$-\sqrt{1-\frac{2k}{1+{k}^{2}}}=-\sqrt{1-\frac{2}{k+\frac{1}{k}}}$．
由圖可知，k≤-1或k≥1．
當(dāng)k≥1時(shí)，k+$\frac{1}{k}$≥2，$0＜\frac{2}{k+\frac{1}{k}}≤1$即$-1≤-\frac{2}{k+\frac{1}{k}}＜0$，∴$0≤1-\frac{2}{k+\frac{1}{k}}＜1$即$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$∈（-1，0]；
當(dāng)k≤-1時(shí)，-k$-\frac{1}{k}$≥2，$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$∈[$-\sqrt{2}$，-1）．
∴$\frac{x-y}{\sqrt{{x}^{2}+{y}^{2}}}$的取值范圍是[$-\sqrt{2}$，-1）∪（-1，0]．
故答案為：[$-\sqrt{2}$，-1）∪（-1，0]．

點(diǎn)評(píng) 本題考查簡(jiǎn)單的線性規(guī)劃，考查了數(shù)形結(jié)合的解題思想方法和數(shù)學(xué)轉(zhuǎn)化思想方法，是難題．