Question

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

• A． $\frac{9}{2}$
• B． $\frac{7}{2}$
• C． 0
• D． -1

Answer 1

分析 由題意得到關(guān)于$\frac{a}，\frac{c}{a}$的不等式組，令x=$\frac{a}$，y=$\frac{c}{a}$換元后作出可行域，進(jìn)一步求得目標(biāo)函數(shù)z=$\frac{2c-b}{a}$=-x+2y的最大值．

解答 解：由$\left\{\begin{array}{l}{a≤b+c}\\{a≥\frac{1}{3}（b+c）}\end{array}\right.$，且$\left\{\begin{array}{l}{b≤a+c}\\{b≥c-2a}\end{array}\right.$，
得$\left\{\begin{array}{l}{\frac{a}+\frac{c}{a}≥1}\\{\frac{a}+\frac{c}{a}≤3}\\{\frac{a}-\frac{c}{a}≤1}\\{\frac{a}-\frac{c}{a}≥-2}\end{array}\right.$，
令x=$\frac{a}$，y=$\frac{c}{a}$，
則$\left\{\begin{array}{l}{x+y≥1}\\{x+y≤3}\\{x-y≤1}\\{x-y≥-2}\end{array}\right.$，z=$\frac{2c-b}{a}$=-x+2y．
作出可行域如圖：
聯(lián)立$\left\{\begin{array}{l}{x-y=-2}\\{x+y=3}\end{array}\right.$，解得A（$\frac{1}{2}，\frac{5}{2}$），
∴z=-x+2y的最大值為$\frac{9}{2}$．
故選：A．

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)合的解題思想方法和數(shù)學(xué)轉(zhuǎn)化思想方法，是中檔題．