GATE 2026 MA – Question 63
(4,0,2,β) is optimal for minimizing x₁+3x₂+2x₃−αx₄, with 4x₁+x₂+x₃=18, −3x₁+2x₃+x₄=2 and all variables nonnegative. For real α,β, find maximum 22(α+β) (integer).
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Correct answer: 206 to 206
Explanation
Feasibility gives β = 2+3(4)−2(2)=10. Eliminate x₃=18−4x₁−x₂ and x₄=11x₁+2x₂−34. Objective becomes $36+34\alpha+(-7-11\alpha)x_1+(1-2\alpha)x_2$. At x₁=4,x₂=0, small positive/negative changes of x₁ are feasible, so optimality forces −7−11α=0: α=−7/11. The remaining x₂ coefficient is 25/11>0, proving global optimality there. Hence 22(10−7/11)=**206**.