Question

12．已知實(shí)數(shù)x、y滿足不等式組$\left\{{\begin{array}{l}{x-2y+1≥0}\\{2x-y-1≤0}\\{4x+2y+1≤0}\\{{x^2}+{y^2}≤1}\end{array}}\right.$，則3x+y的取值范圍為[-$\sqrt{10}$，$-\frac{3}{8}$]．

Answer 1

分析 由約束條件作出可行域，化目標(biāo)函數(shù)為直線方程的斜截式，數(shù)形結(jié)合得到最優(yōu)解，聯(lián)立方程組求得最優(yōu)解的坐標(biāo)，代入目標(biāo)函數(shù)得答案．

解答 解：由約束條件$\left\{{\begin{array}{l}{x-2y+1≥0}\\{2x-y-1≤0}\\{4x+2y+1≤0}\\{{x^2}+{y^2}≤1}\end{array}}\right.$作出可行域如圖，令z=3x+y，
聯(lián)立$\left\{\begin{array}{l}{4x+2y+1=0}\\{2x-y-1=0}\end{array}\right.$，解得A（$\frac{1}{8}$，-$\frac{3}{4}$）此時(shí)z取得最大值，z=$-\frac{3}{8}$．
目標(biāo)函數(shù)與圓相切，可得d=$\frac{|-z|}{\sqrt{9+1}}$=1，解得z=$±\sqrt{10}$，
由圖象可知，z$≥-\sqrt{10}$，
∴3x+y的取值范圍是[-$\sqrt{10}$，$-\frac{3}{8}$]．
故答案為：[-$\sqrt{10}$，$-\frac{3}{8}$]．

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，是中檔題．