Question

13．已知正數(shù)a，b，c滿足約束條件：$\left\{\begin{array}{l}{a≤b+c}\\{a≥\frac{1}{3}（b+c）}\end{array}$且$\left\{\begin{array}{l}{b≤a+c}\\{b≥c-2a}\end{array}$，則$\frac{2c-b}{a}$的最大值為$\frac{9}{2}$．

Answer 1

分析 化簡可得$\left\{\begin{array}{l}{1≤\frac{a}+\frac{c}{a}}\\{1≥\frac{1}{3}（\frac{a}+\frac{c}{a}）}\end{array}\right.$且$\left\{\begin{array}{l}{\frac{a}≤1+\frac{c}{a}}\\{\frac{a}≥\frac{c}{a}-2}\end{array}\right.$，令$\frac{a}$=x，$\frac{c}{a}$=y，從而可得z=$\frac{2c-b}{a}$=2y-x，$\left\{\begin{array}{l}{1≤x+y≤3}\\{-2≤x-y≤1}\end{array}\right.$，從而作圖求解即可．

解答 解：∵a＞0，b＞0，c＞0，$\left\{\begin{array}{l}{a≤b+c}\\{a≥\frac{1}{3}（b+c）}\end{array}$，
∴$\left\{\begin{array}{l}{1≤\frac{a}+\frac{c}{a}}\\{1≥\frac{1}{3}（\frac{a}+\frac{c}{a}）}\end{array}\right.$，
∵$\left\{\begin{array}{l}{b≤a+c}\\{b≥c-2a}\end{array}$，
∴$\left\{\begin{array}{l}{\frac{a}≤1+\frac{c}{a}}\\{\frac{a}≥\frac{c}{a}-2}\end{array}\right.$，
令$\frac{a}$=x，$\frac{c}{a}$=y，（x＞0，y＞0），
則z=$\frac{2c-b}{a}$=2y-x，
$\left\{\begin{array}{l}{\frac{1}{3}（x+y）≤1≤x+y}\\{y-2≤x≤1+y}\end{array}\right.$，
則$\left\{\begin{array}{l}{1≤x+y≤3}\\{-2≤x-y≤1}\end{array}\right.$，
作平面區(qū)域如下，
，
當(dāng)過點(diǎn)B時(shí)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-$\frac{1}{2}$=$\frac{9}{2}$，
故答案為$\frac{9}{2}$．

點(diǎn)評 本題考查了數(shù)形結(jié)合的思想應(yīng)用及轉(zhuǎn)化的思想應(yīng)用，屬于中檔題．