The objective function Z = 4 x 1 + 5 x 2 , subject to 2 x 1 + x 2 ≥ 7 , 2 x 1 + 3 x 2 ≤ 15 , x 2 ≤ 3 , x 1 , x 2 ≥ 0 has minimum value at the point

Step-by-step solution

Value of ( z =4 x ₁+5 x ₂ ) Convert the given inequalities into equalities to get the corner points (2 x₁+x₂=7 ) (i) At (x₁=0, x₂=7 ) and (x₂=0, x₁=3.5 ) So, the corner points of (i) are ((0,7) ) and ((3.5,0) ) (2 x ₁+3 x ₂=15 ) (ii) At ( x ₁=0, x ₂=5 ) and ( x ₂=0, x ₁=7.5 ) So, the corner points of (i) are ((0,5) ) and ((7.5,0) ) ( x ₂=3 )..... (iii) Plot these corner points on the graph paper and the line given in (iii) The shaded part shows the feasible region. At (x₂=3, x₁=2 ) in (i) and (x₁=3 ) in (ii) The co