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Thread: Lagrange finding extreme points.

  1. #1
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    Question Lagrange finding extreme points.

    Hello,
    I've been looking at finding extreme points of the following function f(x,y) = 2x^2 + y^4 =2y^2 - 2xy over the constraint  D = \{(x,y)| x^2 +2y^2 -xy \leq 1\} using Lagrange multipliers.
     \nabla f = \lambda \nabla g, where g is the equation of the constraint.
    This means (4x - 2y, 4y^3 + 4y -2x) = \lambda (2x - y, 2y - x) Following that, I made  4x - 2y = \lambda (2x -y) It's easy to see here that \lambda = 2 or  2x=y Making either substitution into the equation 4y^3 + 4y -2x = \lambda (2y - x), after simplifying, it makes it equal to 2y(y^2 - 1)=0. This implies that y= 0,1 or -1. Substituting these values into the constraint, I find that if y = 0, x = 1 & -1. If y = -1 and 1, values for x won't be real. So this gives me extreme points of (-1,0) and (1,0). Which, in turn, gives me an extreme value of 2 for both points.
    My question is, am I right? Have I found the extreme points or have I missed some? Have I even got the right points?

    Any advice is welcome.
    Thank you for your time.
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  2. #2
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    Quote Originally Posted by Silverflow View Post
    Hello,
    I've been looking at finding extreme points of the following function f(x,y) = 2x^2 + y^4 \mathbin{\color{red}+}2y^2 - 2xy over the constraint  D = \{(x,y)| x^2 +2y^2 -xy \leq 1\} using Lagrange multipliers.
     \nabla f = \lambda \nabla g, where g is the equation of the constraint.
    This means (4x - 2y, 4y^3 + 4y -2x) = \lambda (2x - y, {\color{red}4}y - x) Following that, I made  4x - 2y = \lambda (2x -y) It's easy to see here that \lambda = 2 or  2x=y Making either substitution into the equation 4y^3 + 4y -2x = \lambda (2y - x), after simplifying, it makes it equal to 2y(y^2 - 1)=0. This implies that y= 0,1 or -1. Substituting these values into the constraint, I find that if y = 0, x = 1 & -1. If y = -1 and 1, values for x won't be real. So this gives me extreme points of (-1,0) and (1,0). Which, in turn, gives me an extreme value of 2 for both points.
    My question is, am I right? Have I found the extreme points or have I missed some? Have I even got the right points?
    This looks more or less on the right track, though there are some typos that make the question hard to follow. I assume that the red + and 4 in the above quote are corrections for the = and 2 in the original post.

    Completing the square in the expression for f(x,y) gives f(x,y) = \tfrac12(2x-y)^2 + y^4 + \tfrac32y^2. From this (or by taking partial derivatives), you see that the only critical point of f(x,y) is a global minimum value of 0, at the origin. The origin is inside D, so the minimum value of f(x,y) in D is 0.

    To find the maximum value of f in D, you need to check what happens on the boundary of D, in other words when f satisfies the constraint  x^2 +2y^2 -xy = 1. You do this by the Lagrange multiplier method, and apart from the error of having a 2 that should be 4 you seem to have done this correctly. So the maximum value of f in D is 2, at the points (\pm1,0).
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  3. #3
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    Thank you very much for your reply.
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