五、 给出线性规划问题:(20)Max z=2x1+4x2+x3+x4 x1+3x2+x4≤8 2x1+x2≤6 x2+x3+x4≤6 x1+x2+x3≤9 xj≥0(j=1,2,3,4) 要求:(1)写出其对偶问题;(2)已知原问题最优解为 x1+3x2+x4≤8 2x1+x2≤6 x2+x3+x4≤6 x1+x2+x3≤9 xj≥0(j=1,2,3,4) X x1+3x2+x4≤8 2x1+x2≤6 x2+x3+x4≤6 x1+x2+x3≤9 xj≥0(j=1,2,3,4) =(2,2,4,0),试根据对偶理论,直接求出对偶问题的最优解。

) (x)_(1)+4(x)_(2)-2(x)_(3)+8(x)_(4)=2 -(x)_(1)+2(x)_(2)+3(x)_(3)+4(x)_(4)=1 (x)

线性方程组 ) (x)_(1)+2(x)_(2)-2(x)_(3)=1 2(x)_(1)+4(x)_(2)-4(x)_(3)=2 3(x)_(1)+6(x)_

) (x)_(1)+4(x)_(2)+2(x)_(3)geqslant 8 3(x)_(1)+2(x)_(2)geqslant 6 (x)_(1),(x)_(2

例4 讨论线性方程组-|||- ) (x)_(1)+(x)_(2)+2(x)_(3)+3(x)_(4)=1 (x)_(1)+3(x)_(2)+6(x)_(3)

用克莱姆法则求解线性方程组 } 2x_1+x_2-5x_3+x_4=8x_1-3x_2 -6x_4=9 2x_2-x_3+2x_4=-5x_1+4x_

) 4(x)_(1)+6(x)_(2)geqslant 6 3(x)_(1)+2(x)_(2)geqslant 4 (x)_(1),(x)_(2)geqslan

求二次型 ((x)_(1),(x)_(2),(x)_(3))=4({x)_(2)}^2-3({x)_(3)}^2+4(x)_(1)(x)_(2)-4(x)_(1

3.求解线性方程组 ) (x)_(1)+2(x)_(2)-(x)_(3)+2(x)_(4)=1 2(x)_(1)+4(x)_(2)+(x)_(3)+(x)_(

求线性方程组_(1)+(x)_(2)+(x)_(3)+(x)_(4)=0-|||-_(2)+2(x)_(3)+2(x)_(4)=1-|||-_(1)+2(x)_

用消元法解下列线性方程组:-|||- x1+3x2+5x3-4x4 =1, x1+3x2+2x3-2x4+x5=-1, x1-2x2+x3-x4-x5=3, x