Question

17．已知變量x、y滿足$\left\{\begin{array}{l}{x-y-2≤0}\\{x+2y-5≥0}\\{y-2≤0}\end{array}\right.$，則u=$\frac{3x+y}{x+1}$的取值范圍是[$\frac{5}{2}，\frac{14}{5}$]．

Answer 1

分析 令x+1=m，3x+y=n，代入約束條件后轉(zhuǎn)化關(guān)于m，n的不等式組，數(shù)形結(jié)合得到最優(yōu)解，求出最優(yōu)解的坐標(biāo)，由$\frac{n}{m}$的幾何意義得答案．

解答 解：令x+1=m，3x+y=n，得x=m-1，y=n-3m+3，代入$\left\{\begin{array}{l}{x-y-2≤0}\\{x+2y-5≥0}\\{y-2≤0}\end{array}\right.$，
得$\left\{\begin{array}{l}{4m-n-6≤0}\\{5m-2n≤0}\\{3m-n-1≥0}\end{array}\right.$，作出可行域如圖，

聯(lián)立$\left\{\begin{array}{l}{3m-n-1=0}\\{4m-n-6=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{m=5}\\{n=14}\end{array}\right.$，即C（5，14），
u=$\frac{3x+y}{x+1}$=$\frac{n}{m}$的幾何意義為可行域內(nèi)的動(dòng)點(diǎn)與原點(diǎn)連線的斜率，
∵${k}_{OA}={k}_{AB}=\frac{5}{2}$，${k}_{OC}=\frac{14}{5}$，
∴u=$\frac{3x+y}{x+1}$的取值范圍是[$\frac{5}{2}，\frac{14}{5}$]．
故答案為：[$\frac{5}{2}，\frac{14}{5}$]．

點(diǎn)評(píng) 本題考查了簡(jiǎn)單的線性規(guī)劃，考查了數(shù)形結(jié)合及數(shù)學(xué)轉(zhuǎn)化思想方法，是中檔題．