For the L.P. problem Maxz = 3x1 + 2x2 such that 2x1 - x2 ≥ 2 , x1 + 2x2 ≤ 8 and x1 , x2 ≥ 0, z -

Question

For the L.P. problem Maxz = 3x1 + 2x2 such that 2x1 - x2 ≥ 2 , x1 + 2x2 ≤ 8 and x1 , x2 ≥ 0, z -  (A) 12 (B) 24 (C) 36 (D) 40

Tardigrade Admin · Accepted Answer

Change the inequalities into equations and draw the graph of lines, thus we get the required feasible region. It is a bounded region, bounded by the vertices A(1,0),B(8,0) and ( (12/5) . (14/5) ) . Now by evaluation of the objective function for the vertices of feasible region it is found to be maximum at (8,0). Hence the solution is z - 3 × 8 + 0 × 2 - 24 .

Q. For the L.P. problem Maxz​=3x1​+2x2​ such that 2x1​−x2​≥2,x1​+2x2​≤8 and x1​,x2​≥0,z−

12

24

36

40

Q. For the L.P. problem Max $_{z} = 3 x_{1} + 2 x_{2}$ such that $2 x_{1} - x_{2} \geq 2, x_{1} + 2 x_{2} \leq 8$ and $x_{1}, x_{2} \geq 0, z -$