http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 16 Jul 2014 22:34:36 -0500http://calc3.askbot.com/question/101/evaluating-the-maxmin-using-lagrange-method-on-the-turn-in-homework/In class our usual problem might state that we are looking to see whether we have a max or min at and intersection between two functions, and to solve this we use the lagrange method by setting both gradients equal to each other times a constant $\nabla f = \nabla \lambda g$, where $g(x,y)$ is equal to a constant. In this problem, we are given two constraint functions $g(x,y)$ and $h(x,y)$ that are both equal to $x$. Do we simply bring $x$ over to either side of the inequality equation and set that equal to 0 to create the "third" equation to solve our system? Or is there something else that I'm missing? $$ f(x,y) = x^3 - 2xy +y^2 -3x $$ constrained by: $$ y^2-4 \leq x \leq 1-y $$ As a follow up, I solved my system of equations such that the third equation looked like so: $$ 1-y=x $$ But as I solved this system for \lambda , $x$ and $y$, I ended up getting an imaginary root. looking at the pictures people have posted, I don't think that this should be the case. Graphing this function in matlab yielded this image (not sure how to get the constraints on it too)  And the corresponding contour plot to go with it Wed, 16 Jul 2014 01:42:44 -0500http://calc3.askbot.com/question/101/evaluating-the-maxmin-using-lagrange-method-on-the-turn-in-homework/http://calc3.askbot.com/question/101/evaluating-the-maxmin-using-lagrange-method-on-the-turn-in-homework/?answer=115#post-id-115What I did was to bring over the $x$ to one side to create my "$g$" functions for the Lagrange multiplier method, but then just used the equality version of that side of the inequality to get the third equation. At first, I didn't move over $x$ and this caused me problems when trying to find the extrema. My guess is that the equation represents a contour line of a function of both $x$ and $y$, so if we don't move them over to create $g$, we aren't actually getting the right function. To get the third equation, I think that because we are essentially checking the "border" of the domain, we are implicitly assuming that the equality version of that inequality is true, so we can definitely use it as an equation. I also got complex roots for one of my systems of equations; I think it makes sense that some of our equations would have solutions that happened to be complex; I'm not sure of any geometric reason this shouldn't happen. I did not find any systems that had *no* real solutions, though. Wed, 16 Jul 2014 22:34:36 -0500http://calc3.askbot.com/question/101/evaluating-the-maxmin-using-lagrange-method-on-the-turn-in-homework/?answer=115#post-id-115