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Author Question: A linear programming maximization problem has been solved. In the optimal solution, two resources ... (Read 136 times)

sammy

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on: Dec 3, 2019

Question 1

Consider the following constraints from a two-variable linear program.
(1) X ≥ 1
(2) Y ≥ 1
(3) X + Y ≤ 9
If these are the only constraints, which of the following points (X, Y) cannot be the optimal solution?
◦ (1, 1)
◦ (4, 4)
◦ (1, 8)
◦ (8, 1)
◦ The question cannot be answered without knowing the objective function.

Question 2

A linear programming maximization problem has been solved. In the optimal solution, two resources are scarce. If an added amount could be found for only one of these resources, how would the optimal solution be changed?
◦ The solution stays the same; the extra resource can't be used without more of the other scarce resource.
◦ The extra resource will cause the value of the objective to rise.
◦ The optimal mix will be rearranged to use the added resource, and the value of the objective function will rise.
◦ The shadow price of the added resource will rise.
◦ The extra resource will cause the value of the objective to fall.
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Marked as best answer by sammy on Dec 3, 2019

polinasid

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Reply #1 on: Dec 3, 2019
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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sammy

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Reply 2 on: Dec 3, 2019
:D TYSM

CAPTAINAMERICA

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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