In the LP with Min Z = 26X1 + 14X2 + 12X3 subject to 2X1 + 3X2 + X3 ≥ 30, 6X1 + 3X2 + X3 ≥ 50, X1, X2, X3 ≥ 0, what is the optimal value of X3?

• A. 1
• B. 2
• C. 3
• D. 0

Answer: (D)

Minimizing with constraints of the form ≥ often yields a solution at a corner where the active constraints bind. Here, set both inequalities to equalities to locate that corner and see how the objective behaves there.

2X1 + 3X2 + X3 = 30

6X1 + 3X2 + X3 = 50

Subtracting the first from the second gives 4X1 = 20, so X1 = 5. Plugging back into the first gives 2(5) + 3X2 + X3 = 30, so 3X2 + X3 = 20 and X3 = 20 − 3X2.

Feasibility requires X2 ≥ 0 and X3 ≥ 0, which means 0 ≤ X2 ≤ 20/3. The objective is Z = 26X1 + 14X2 + 12X3 = 26(5) + 14X2 + 12(20 − 3X2) = 130 + 14X2 + 240 − 36X2 = 370 − 22X2. So Z decreases as X2 increases along this line; the minimum occurs at the largest X2 allowed, X2 = 20/3, which makes X3 = 0.

At this point both constraints are tight: 2(5) + 3(20/3) + 0 = 30 and 6(5) + 3(20/3) + 0 = 50. Therefore the optimal value of X3 is 0.