{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mthsc206-fall-2010-notes-15.8

mthsc206-fall-2010-notes-15.8 - g x y z = k 2 Evaluate f at...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
15.8 Lagrange Multipliers Theorem: Let f and g be functions of 2( or 3) variables with continuous first partial derivatives on some open set containing the constraint curve (or surface) g ( x, y ) = k (or g ( x, y, z ) = k ) and assume g = 0 at any point on this curve (or surface). If f has a constrained relative extremum, then this extremum occurs at a point P 0 on the constraint curve (or surface) at which the gradient vectors f ( P 0 ) and g ( P 0 ) are parallel; i.e., there is some number λ such that f ( P 0 ) = λ g ( P 0 ) . Method of Lagrange Multipliers: To find the maximum and minimum values of f ( x, y, z ) subject to the constraint g ( x, y, z ) = k (assuming that these extreme values exist and g = 0 on the surface g ( x, y, z ) = k ): 1. Find all values of x , y , z , and λ such that f ( x, y, z ) = λ g ( x, y, z ) AND
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g ( x, y, z ) = k. 2. Evaluate f at all of the points ( x, y, z ) that result from step 1. The largest of these values is the maximum of f subject to the constraint; the smallest of these values is the minimum of f subject to the constraint. Example 15.8.1 Redo fnding the absolute extrema oF f ( x, y ) = x 2 + 2 y 2-x on the region x 2 + y 2 ≤ 4 using the Method oF Lagrange Multipliers. Example 15.8.2 ±ind the extrema oF f ( x, y ) = xy iF ( x, y ) is restricted to the ellipse 4 x 2 + y 2 = 4 . Example 15.8.3 IF f ( x, y, z ) = 4 x 2 + y 2 + 5 z 2 , fnd the point on the plane 2 x + 3 y + 4 z = 12 at which f ( x, y, z ) has its least value. 33...
View Full 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