Question

13．已知正数a，b，c满足约束条件：$\left\{\begin{array}{l}{a≤b+c}\\{a≥\frac{1}{3}（b+c）}\end{array}$且$\left\{\begin{array}{l}{b≤a+c}\\{b≥c-2a}\end{array}$，则$\frac{2c-b}{a}$的最大值为$\frac{9}{2}$．

Answer 1

分析 化简可得$\left\{\begin{array}{l}{1≤\frac{b}{a}+\frac{c}{a}}\\{1≥\frac{1}{3}（\frac{b}{a}+\frac{c}{a}）}\end{array}\right.$且$\left\{\begin{array}{l}{\frac{b}{a}≤1+\frac{c}{a}}\\{\frac{b}{a}≥\frac{c}{a}-2}\end{array}\right.$，令$\frac{b}{a}$=x，$\frac{c}{a}$=y，从而可得z=$\frac{2c-b}{a}$=2y-x，$\left\{\begin{array}{l}{1≤x+y≤3}\\{-2≤x-y≤1}\end{array}\right.$，从而作图求解即可．

解答 解：∵a＞0，b＞0，c＞0，$\left\{\begin{array}{l}{a≤b+c}\\{a≥\frac{1}{3}（b+c）}\end{array}$，
∴$\left\{\begin{array}{l}{1≤\frac{b}{a}+\frac{c}{a}}\\{1≥\frac{1}{3}（\frac{b}{a}+\frac{c}{a}）}\end{array}\right.$，
∵$\left\{\begin{array}{l}{b≤a+c}\\{b≥c-2a}\end{array}$，
∴$\left\{\begin{array}{l}{\frac{b}{a}≤1+\frac{c}{a}}\\{\frac{b}{a}≥\frac{c}{a}-2}\end{array}\right.$，
令$\frac{b}{a}$=x，$\frac{c}{a}$=y，（x＞0，y＞0），
则z=$\frac{2c-b}{a}$=2y-x，
$\left\{\begin{array}{l}{\frac{1}{3}（x+y）≤1≤x+y}\\{y-2≤x≤1+y}\end{array}\right.$，
则$\left\{\begin{array}{l}{1≤x+y≤3}\\{-2≤x-y≤1}\end{array}\right.$，
作平面区域如下，
，
当过点B时有最大值，
由$\left\{\begin{array}{l}{x=y-2}\\{x=3-y}\end{array}\right.$解得，x=$\frac{1}{2}$，y=$\frac{5}{2}$，
故$\frac{2c-b}{a}$=2y-x=5-$\frac{1}{2}$=$\frac{9}{2}$，
故答案为$\frac{9}{2}$．

点评 本题考查了数形结合的思想应用及转化的思想应用，属于中档题．