分别用图解法和单纯形法求解下列线性规划 max z =2x1+x2 {3x1+5x2 ≤15 {6x1+2x2 ≤24 {x1 ,x2 ≥ 0
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![分别用图解法和单纯形法求解下列线性规划 max z =2x1+x2 {3x1+5x2 ≤15 {6x1+2x2 ≤24 {x1 ,x2 ≥ 0](/uploads/image/z/4480253-53-3.jpg?t=%E5%88%86%E5%88%AB%E7%94%A8%E5%9B%BE%E8%A7%A3%E6%B3%95%E5%92%8C%E5%8D%95%E7%BA%AF%E5%BD%A2%E6%B3%95%E6%B1%82%E8%A7%A3%E4%B8%8B%E5%88%97%E7%BA%BF%E6%80%A7%E8%A7%84%E5%88%92+max+z+%3D2x1%2Bx2+%EF%BD%9B3x1%2B5x2+%E2%89%A415+%EF%BD%9B6x1%2B2x2+%E2%89%A424+%EF%BD%9Bx1+%2Cx2+%E2%89%A5+0)
分别用图解法和单纯形法求解下列线性规划 max z =2x1+x2 {3x1+5x2 ≤15 {6x1+2x2 ≤24 {x1 ,x2 ≥ 0
分别用图解法和单纯形法求解下列线性规划 max z =2x1+x2 {3x1+5x2 ≤15 {6x1+2x2 ≤24 {x1 ,x2 ≥ 0
分别用图解法和单纯形法求解下列线性规划 max z =2x1+x2 {3x1+5x2 ≤15 {6x1+2x2 ≤24 {x1 ,x2 ≥ 0
才2个未知数,图解法自己画图.
单纯形:
标准型:maxz=2X1+X2+0X3+0X4
ST: 3X1+5X2+X3=15
6X1+2X2+X4=24
Cj→ 2 1 0 0
Cb 基 b X1 X2 X3 X4
0 X3 15 3 5 1 0
0 X4 24 [6] 2 0 1
检验数 2 1 0 0
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0 X3 3 0 [4] 1 -1/2
2 X1 4 1 1/3 0 1/6
检验数 0 1/3 0 -1/3
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1 X2 3/4 0 1 1/4 -1/8
2 X1 2/9 1 0 -1/12 145/24
检验数 0 0 -1/12 -17/36
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所以X=(2/9 3/4 0 0)
maxz=43/36