Question

7．如果實(shí)數(shù)x，y滿足約束條件$\left\{\begin{array}{l}x+2y-4≥0\\ x-y+2≥0\\ 2x+y-3≤0\end{array}\right.$，則（x+2）²+y²的最小值為8．

Answer 1

分析 由約束條件作出可行域，由（x+2）²+y²的幾何意義，即原點(diǎn)O（-2，0）到直線3x+4y-5=0的距離求得答案．

解答 解：由約束條件$\left\{\begin{array}{l}x+2y-4≥0\\ x-y+2≥0\\ 2x+y-3≤0\end{array}\right.$作出可行域如圖，

由圖可知，z=x²+y²的最小值為原點(diǎn)O（-2，0）到$\left\{\begin{array}{l}{x+2y-4=0}\\{x-y+2=0}\end{array}\right.$的交點(diǎn)的距離的平方，
（0+2）²+2²=8．
故答案為：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．