# Homework Help: Lagrange multiplier problem - function of two variables with one constraint

1. Mar 10, 2012

### abery

1. The problem statement, all variables and given/known data
Find the maximum and minimum values of f(x,y) = 2x^2+4y^2 - 4xy -4x
on the circle defined by x^2+y^2 = 16.

2. Relevant equations
Lagrange's method, where f_x = lambda*g_x, f_y= lambda*g_y (where f is the given function and g(x,y) is the circle on which we are looking for the extrema)

3. The attempt at a solution
Computed the partials, and was able to end up with an equation like 2y -2 = $\lambda$*(x-y)

From this, critical points look like they might be (1, root 15) and (root 15, 1) but this does not seem to be the answer.

Any help at all is much appreciated!

2. Mar 10, 2012

### HallsofIvy

Well, it would have helped if you had show precisely what you did, not just generalities. "end up with an equation like 2y -2 = λ *(x-y)". Actually, I think you have f and g reversed but it really doesn't matter whether you use $f_x= \lambda g_x$ or $g_x=(1/\lambda) f_x$

$f(x,y)= 2x^2+ 4y^2- 4xy- 4x$ and $g(x,y)= x^2+ y^2= 16$.
so $f_x= 4x- 4y- 4$ and $g_x= 2x$. Your first equation is $4x- 4y- 4= 2\lambda x$. $f_y= 8y- 4x$ and $g_y= 2y$. Your second equation is $8y- 4x= 2\lambda y$. Those, together with the condition that $x^2+ y^2= 16$, give you three equations to solve for x, y, and $\lambda$.

But the value of $\lambda$ is not really necessary to solve this problem and I find that it is often best to eliminate $\lambda$ from the first two equations by dividing one by the other.

3. Mar 10, 2012

### abery

Thanks for your help, HallsOfIvy. My problem is with solving the resulting system of equations. (Thanks for the suggestion to divide the equations)

After doing that and rearranging the first two, I get: x2 - xy -y2 - y =0

And using the third (original) constraint, I get -2y2 - xy - y - 16 = 0, but am not sure how to proceed.