UNIT8 - Unit 8 Lagranges Method and Calculus of...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Unit 8 Lagrange’s Method and Calculus of Trigonometric Functions Constrained Optimization: Lagrange’s Method Old Problem: Minimize z = x 2 + y 2 +2 subject to the constraint x + y = 8. Last time we solved a similar problem by using the constraint to eliminate a variable. i.e. the problem becomes: Minimize: z = x 2 + (8 x ) 2 + 2 Now easy to solve: x = 4 y = 4 , z = 4 New Problem: Minimize f ( x, y, z ) = xyz subject to the constraint xy + 2 xz + 4 yz = 216. The old method does not work here since we can not use the constraint to solve for one of the variables, thus we require a new method. To this end we turn to a method developed by Joseph Lagrange in the 1700’s. Lagrange’s Method For the problem of optimizing z = f ( x, y ) subject to the constraint g ( x, y ) = 0, let 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
F ( x, y, λ ) = f ( x, y ) + λg ( x, y ) The local extreme points ( x, y ) will be found in the set of points ( x, y, λ ) that satisfy the three equations: F x ( x, y, λ ) = 0 F y ( x, y, λ ) = 0 F λ ( x, y, λ ) = 0 This process generalizes to the case where we have f ( x, y, z ). Example 1. Minimize f ( x, y ) = x 2 + y 2 +2 subject to g ( x, y ) = x + y 8 = 0. Using Lagrange’s Method, F ( x, y, λ ) = ( x 2 + y 2 + 2 ) + λ ( x + y 8) F x = 2 x + λ = 0 λ = 2 x (1) F y = 2 y + λ = 0 λ = 2 y (2) F λ = x + y 8 = 0 (3) Setting (1)=(2) we get 2 x = 2 y x = y Substituting this into (3) yields: x + x 8 = 0 x = 4 And since we have x = y , we get y = 4 as well. So our solution is ( x, y ) = (4 , 4).
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern