Lagrange Multipliers with ellipse

[Quincy]

Homework Statement

Find the points on the ellipse x² + 2y² = 1 where f(x,y) = xy has its extreme values.

Homework Equations

The Attempt at a Solution

f(x,y,z) = x² + y² + z² -- constraint
g(x,y,z) = x² + 2y² -1 = 0

gradient of f = \lambda * gradient of g

2xi + 2yj + 2zk = \lambda2xi + \lambda4yj

2x = \lambda2x
2y = \lambda4y
2z = 0

I don't know where to go from here, can someone help me out?

 

[Quincy]

Sorry, I made a mistake in my original post, it should be:

f(x,y,z) = xy
g(x,y,z) = x² + 2y² - 1

yi + xj = \lambda(2xi + 4yj)
y = 2x\lambda
x = 4y\lambda

y = 2(4y\lambda)\lambda = 8y\lambda²

\lambda = sqrt(1/8)

where do i go from here?

 

[Office_Shredder]

you have three equations and three unknowns... the two gradient equations and the fact that x²+2y²=1. Now that you have solved for one unknown, can you make it two equations with two unknowns?

 

[Quincy]

How about this one:

Find the minimum distance from the surface x² - y² - z² = 1 to the origin. (function being minimized = x² + y² + z²)

2xi + 2yj + 2zk = \lambda(2xi - 2yj -2zk)

2x = 2x\lambda
2y = -2y\lambda
2z = -2z\lambda
x² = y² +z² + 1

- I can't figure out how to solve this system of equations, any tips?

 

[HallsofIvy]

These are pretty close to being trivial. From the first equation, if x\ne 0, \lambda must be 1. From the second equation, if y\ne 0, \lambda must be -1, and from the third, if z\ne 0, \lambda must be -1.

Since \lambda cannot be both 1 and -1, at least one of the coordinates must be 0!

Suppose x= 0. Then the constraint becomes -y^2- z^2= 1 which is impossible. If x is not 0, then \lambda must be 1, not -1, so y and z must be 0. The constraint becomes x^2= 1.

Looking at it geometrically gives an easy check. This is a "hyperboloid of two sheets" with the x-axis as axis of symmetry.