题目:若x,y满足x≥0,y≥0,2x+y≤6,x+2y≤6,,求z=2x+3y的最大值·分析1变换角度,建立zoy坐标系,通过观察可行域中的横坐标z的取值范围,从而得到z的最值·解法1由z=2x+3y得x=(z-3y)/2,则(z-3y)/2≥0,y≥0,z-3y+y≤6,(z-3y)/2+2y≤6,即z-3y≥0,y≥0,z-2y≤6,z+y≤12,作出可行域如 Title: If x,y satisfies x≥0, y≥0, 2x+y≤6, and x+2y≤6, find the maximum of z=2x+3y. Analyze the 1 transformation angle and establish the zoy coordinate system. The range of values ​​of the abscissa z in the feasible region, so that the value of z is obtained. Solution 1 is obtained from z=2x+3y where x=(z-3y)/2, then (z-3y)/2≥0,y ≥ 0, z-3y + y ≤ 6, (z-3y) / 2+2y ≤ 6, ie, z-3y ≥ 0, y ≥ 0, z-2y ≤ 6, z + y ≤ 12, making feasible regions such as