Max/Min of x2−2xy+7y2 on Ellipse x2+4y2=1 w/ Lagrange Multiplier

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Homework Statement


Use the Lagrange Multiplier method to find the maximum and minimum values of x2 − 2xy + 7y2 on the ellipse x2 + 4y2 = 1.

Homework Equations


Lagrange multiplier method

The Attempt at a Solution


L(x,y,z,λ) = x2 − 2xy + 7y2 - λ(x2 + 4y2 - 1)
Find Lx, Ly, Lλ
Then, solve for x and y?
 
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Where did you get the L(x,y,z,λ)? You don't need the z in there because we're dealing with functions of x and y, not functions of x, y, and z.

Also, λ is a constant: you don't need to solve for Lλ. You don't need to solve for Lz either because there's no such variable as z. Just set Lx=0 and Ly=0 and solve. Don't forget to satisfy the initial constraint of x^2+4y^2=1!
 
Just an added note: some people are taught to find L_\lambda. Because if we are trying to extremize F(x) subject to the constraint G(x)= 0, we look at L= F(x)+ \lambda G(x), L_{\lamba}= G(x)= 0 is just the constraint again.
 
@ideasrule: Sorry, ignore the "z."
@HallsofIvy: Thanks. And yes, that's how I learned it.

Also, once I find the critical points, do I just plug them back into the original function I want to extremize to see if it's a max or min?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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