Evaluating the max/min using Lagrange method on the turn-in homework
asked 2014-07-16 01:42:44 -0600
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
