Question

11．已知實(shí)數(shù)x，y滿足$\left\{\begin{array}{l}{x-y+2≥0}\\{x+y-4≥0}\\{4x-y-4≤0}\end{array}\right.$，則z=3x-y的取值范圍為（　�。�

• A． [0，$\frac{12}{5}$]
• B． [0，2]
• C． [2，$\frac{12}{5}$]
• D． [2，$\frac{8}{3}$]

Answer 1

分析 先畫出可行域，再把目標(biāo)函數(shù)變形為直線的斜截式，根據(jù)其在y軸上的截距即可求之．

解答 解：畫出$\left\{\begin{array}{l}{x-y+2≥0}\\{x+y-4≥0}\\{4x-y-4≤0}\end{array}\right.$的可行域，如圖所示
$\left\{\begin{array}{l}{x-y+2=0}\\{x+y-4=0}\end{array}\right.$解得A（1，3）、由$\left\{\begin{array}{l}{x+y-4=0}\\{4x-y-4=0}\end{array}\right.$解得B（$\frac{8}{5}$，$\frac{12}{5}$），
把z=3x-y變形為y=3x-z，則直線經(jīng)過點(diǎn)A時(shí)z取得最小值；經(jīng)過點(diǎn)B時(shí)z取得最大值．
所以z_min=3×1-3=0，z_max=3×$\frac{8}{5}$-$\frac{12}{5}$=$\frac{12}{5}$．
即z的取值范圍是[0，$\frac{12}{5}$]．
故選：A．

點(diǎn)評(píng) 本題考查利用線性規(guī)劃求函數(shù)的最值．利用數(shù)形結(jié)合是解決線性規(guī)劃題目的常用方法．