Question

4．設(shè)變量x，y滿足約束條件$\left\{\begin{array}{l}{x-y-1≤0}\\{x+y≥0}\\{x+2y-4≥0}\end{array}\right.$，則z=x-2y的最大值為$-\frac{1}{2}$．

Answer 1

分析 由題意作平面區(qū)域，化簡z=3x-2y為y=$\frac{1}{2}$x-$\frac{1}{2}z$，從而可得-$\frac{1}{2}z$是直線y=$\frac{1}{2}$x-$\frac{1}{2}z$的截距，從而解得．

解答 解：由題意作變量x，y滿足約束條件$\left\{\begin{array}{l}{x-y-1≤0}\\{x+y≥0}\\{x+2y-4≥0}\end{array}\right.$平面區(qū)域如圖，
化簡z=x-2y為y=$\frac{1}{2}$x-$\frac{1}{2}z$，-$\frac{z}{2}$是直線y=$\frac{1}{2}$x-$\frac{1}{2}z$的截距，
由$\left\{\begin{array}{l}{x+y=0}\\{x-y-1=0}\end{array}\right.$，解得A（$\frac{1}{2}$，-$\frac{1}{2}$）
故過點(diǎn)A（$\frac{1}{2}，-\frac{1}{2}$）時，
z=x-2y有最大值為$\frac{1}{2}$-2×$\frac{1}{2}$=-$\frac{1}{2}$，
故答案為：-$\frac{1}{2}$．

點(diǎn)評 本題考查了線性規(guī)劃的解法及數(shù)形結(jié)合的思想應(yīng)用．