Revision history [back]

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.

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.

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)