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Author Question: A maximizing linear programming problem has two constraints: 2X + 4Y (Read 117 times)

washai

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on: Dec 3, 2019

Question 1

What combination of a and b will yield the optimum for this problem?
Maximize $6a + $15b, subject to (1) 4a + 2b < 12 and (2) 5a + 2b < 20 and (3) a, b ≥ 0.
◦ a = 0, b = 0
◦ a = 0, b = 6
◦ a = 3, b = 3
◦ a = 6, b = 0
◦ cannot solve without values for a and b

Question 2

A maximizing linear programming problem has two constraints: 2X + 4Y < 100 and 3X + 10Y < 210, in addition to constraints stating that both X and Y must be nonnegative. The corner points of the feasible region of this problem are
◦ (20, 15).
◦ (0, 0), (70, 0), (25, 0), and (15, 20).
◦ (0, 0), (50, 0), (0, 21), and (20, 15).
◦ (0, 0), (0, 100), and (210, 0).
◦ none of the above.
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Marked as best answer by washai on Dec 3, 2019

epscape

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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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washai

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Reply 2 on: Dec 3, 2019
:D TYSM

upturnedfurball

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

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