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Questions › JNU › JNU SIS 2012 › Q10
4 marks, −1Mathematics › Constrained optimisation and linear programming › Equality-constrained optimisation and Lagrange multipliersModerateabout 2 min

JNU SIS 2012 Question 10: Constrained optimisation and linear programming

Minimize y=x1+x2y = x_1 + x_2y=x1​+x2​ subject to 1−x11/2−x2=01 - x_1^{1/2} - x_2 = 01−x11/2​−x2​=0

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Minimize y = x₁ + x₂ subject to 1 - x₁^(1/2) - x₂ = 0 (A) x₁* = 1/4, x₂* = 1/2 (B) x₁* = 1/4, x₂* = 1 (C) x₁* = 4, x₂* = 2 (D) x₁* = 1/2, x₂* = 1/4

Also written as: Minimize y = x_1 + x_2 subject to 1 - x_1^(1/2) - x_2 = 0

Constrained optimisation and linear programming has come up 44 times in 7 exams since 2007.

What this question tests

This question is on Equality-constrained optimisation and Lagrange multipliers, part of Constrained optimisation and linear programming in Mathematics.

Optimising a function subject to equality constraints, and what the multiplier means.

TestStop has 14 past-paper questions on this subtopic.

Practise Constrained optimisation and linear programmingMore on Equality-constrained optimisation and Lagrange multipliers

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