Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
maxz₁ +37₂ s.t. +2²=10

maxz₁ +37₂ s.t. +2²=10

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

how to solve the systems of equations please teach  explain step by step also  Verify if max or min please solve the hessians the other hessians 

Lagrangian: The FOC becomes: . s.t. 21+22=10 maxz₁ +373 ZER² L(z; λ) = (₁ +37₂)-A₁(x+x-10) ƏL(x; X) Əf(x) Əgi(x) ?х ?х -A₁ əzi ƏL(x; X) əx₂ ƏL(x; X) ƏX₁ =1-2₁₁0 21 = af (1) ₁² Əgi(z) Əz₂ 01₂ =3-2₁₂01 =+= 10 We obtain two solutions to this system of equations (21, 22, A1) = (1, 3, ¹/2) . (₁, 2, ₁) = (-1, -3, -1/2) Let's look at these two candidates for a maximum: • For X = ¹/2 the Lagrangian is: L(x, X*) = x₂ + 3x₂ − ²½ (2² + x²³) – -5 This function is concave ⇒ (1,72) = (1,3) is a maximum point. H(x) = One can either see this from the equation, or one can check the Hessian matrix: -(¹9) 1 BH = 3 20₁ This matrix is negative semi-definite: principle minors of order k = 1 are negative (-1), principle minor of order k = 2 is |H(z)| = 1. Alternatively one can check this via the bordered Hessian here: 0 2 6 2 -1 0 6 0 -1 ⇒ |BH| = 40> 0 ⇒ x* = (1, 3) is a maximum.
Transcribed Image Text:Lagrangian: The FOC becomes: . s.t. 21+22=10 maxz₁ +373 ZER² L(z; λ) = (₁ +37₂)-A₁(x+x-10) ƏL(x; X) Əf(x) Əgi(x) ?х ?х -A₁ əzi ƏL(x; X) əx₂ ƏL(x; X) ƏX₁ =1-2₁₁0 21 = af (1) ₁² Əgi(z) Əz₂ 01₂ =3-2₁₂01 =+= 10 We obtain two solutions to this system of equations (21, 22, A1) = (1, 3, ¹/2) . (₁, 2, ₁) = (-1, -3, -1/2) Let's look at these two candidates for a maximum: • For X = ¹/2 the Lagrangian is: L(x, X*) = x₂ + 3x₂ − ²½ (2² + x²³) – -5 This function is concave ⇒ (1,72) = (1,3) is a maximum point. H(x) = One can either see this from the equation, or one can check the Hessian matrix: -(¹9) 1 BH = 3 20₁ This matrix is negative semi-definite: principle minors of order k = 1 are negative (-1), principle minor of order k = 2 is |H(z)| = 1. Alternatively one can check this via the bordered Hessian here: 0 2 6 2 -1 0 6 0 -1 ⇒ |BH| = 40> 0 ⇒ x* = (1, 3) is a maximum.
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS