实数x,y满足x-y-1≦0,x+y-3≦0,x≧1,则目标函数z=2x-y的最大值为 4.3.0.-1
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![实数x,y满足x-y-1≦0,x+y-3≦0,x≧1,则目标函数z=2x-y的最大值为 4.3.0.-1](/uploads/image/z/5264282-2-2.jpg?t=%E5%AE%9E%E6%95%B0x%2Cy%E6%BB%A1%E8%B6%B3x-y-1%E2%89%A60%2Cx%2By-3%E2%89%A60%2Cx%E2%89%A71%2C%E5%88%99%E7%9B%AE%E6%A0%87%E5%87%BD%E6%95%B0z%3D2x-y%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA+4.3.0.-1)
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实数x,y满足x-y-1≦0,x+y-3≦0,x≧1,则目标函数z=2x-y的最大值为 4.3.0.-1
实数x,y满足x-y-1≦0,x+y-3≦0,x≧1,则目标函数z=2x-y的最大值为 4.3.0.-1
实数x,y满足x-y-1≦0,x+y-3≦0,x≧1,则目标函数z=2x-y的最大值为 4.3.0.-1
最终答案:4
x-y-1≤0
x+y-3≤0
两式相加,2x≤4
又∵x≥1
∴1≤x≤2
两式相减,y≤2
∵x-y-1≤0
∴y≥x-1
而x≥1
∴y≥0
∴0≤y≤2
所以0≤2x-y≤4(把x的最大值2,y的最小值0代入2x-y中,就得出最大值是4)
最大值为4
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