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Consider maximising 10 - x2 - 4y2 + 2xy where where x > 12. What y-value solves this maximisation problem?

Consider maximising 10 - x2 - 4y2 + 2xy where where x > 12. What y-value solves this maximisation problem?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.2: Transformations Of Quadratic Functions
Problem 49PFA
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Consider maximising 10 - x2 - 4y2 + 2xy where where x > 12. What y-value solves this maximisation problem? Your Answer: Answer Hide hint for Question 1 First, write the Lagrangian. Remember to rewrite the inequality constraint in the form 0 ... beforehand. You should get two first order conditions as usual from it: Ly = 0 and Ly = 0. For the third equation, using the complementary slackness condition. There are two cases: 1. constraint holds with strict equality and the multiplier is non-negative 2. constraint holds with strict inequality and the multiplier is zero Each case should come with one equation and one check to perform on the potential critical points. Find all potential critical points in every case. If there is more than one critical point, then compare them to find the maximum.
Transcribed Image Text:Consider maximising 10 - x2 - 4y2 + 2xy where where x > 12. What y-value solves this maximisation problem? Your Answer: Answer Hide hint for Question 1 First, write the Lagrangian. Remember to rewrite the inequality constraint in the form 0 ... beforehand. You should get two first order conditions as usual from it: Ly = 0 and Ly = 0. For the third equation, using the complementary slackness condition. There are two cases: 1. constraint holds with strict equality and the multiplier is non-negative 2. constraint holds with strict inequality and the multiplier is zero Each case should come with one equation and one check to perform on the potential critical points. Find all potential critical points in every case. If there is more than one critical point, then compare them to find the maximum.
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