The optimal solution of the linear programming problem is at the intersection of constraints 1 and 2.

Max 2x1 + x2

s.t. 4x1 + 1x2 4x1 + 3x2 1x1 + 2x2 £ 300

x1 , x2 > 0

Compute the dual prices for the three constraints.

.45, .25, 0

.25, .25, 0

0, .25, .45

.45, .25, .25

Max 2x1 + x2

s.t. 4x1 + 1x2 4x1 + 3x2 1x1 + 2x2 £ 300

x1 , x2 > 0

Compute the dual prices for the three constraints.

.45, .25, 0

.25, .25, 0

0, .25, .45

.45, .25, .25

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