How to Solve LaGrange Multiplier Problems for Intersection of Surfaces?

[Baumer8993]

Homework Statement

Consider the intersection of the elliptic paraboloid Z = X²+4Y² , and the cylinder X²+Y²= 1. Use Lagrange multipliers to find the highest, and lowest points on the curve of intersection.

Homework Equations

The gradient equations of both functions.

The Attempt at a Solution

I have ∇f= <2X, 8Y> and ∇g= <2X, 2Y>. the constraint equation is X²+Y² = 1.

I set the equations equal to get:

2X = (2X)λ 8Y = (2Y)λ

When I try to solve it always removes the variable. Where do I go from here to solve?

 

[Ray Vickson]

Homework Statement

Consider the intersection of the elliptic paraboloid Z = X²+4Y² , and the cylinder X²+Y²= 1. Use Lagrange multipliers to find the highest, and lowest points on the curve of intersection.

Homework Equations

The gradient equations of both functions.

The Attempt at a Solution

I have ∇f= <2X, 8Y> and ∇g= <2X, 2Y>. the constraint equation is X²+Y² = 1.

I set the equations equal to get:

2X = (2X)λ 8Y = (2Y)λ

When I try to solve it always removes the variable. Where do I go from here to solve?

Look at the first equation $2x = 2x \lambda$. Can you cancel the $2x$ on both sides? Why, or why not?

 

[Baumer8993]

I think you can, but it would just leave me with λ = 1.

 

[Ray Vickson]

I think you can, but it would just leave me with λ = 1.

OK, so? Work it through to the end.

BTW: there is another possibility having λ ≠ 1; can you see why?

 

[Baumer8993]

I see that λ could also equal 4, but where do I go from plugging in the lambda values? I just end up with 2x=2x, and 8y=2Y, or do I need to use both λ lambda values at the same time?

 

[Ray Vickson]

I see that λ could also equal 4, but where do I go from plugging in the lambda values? I just end up with 2x=2x, and 8y=2Y, or do I need to use both λ lambda values at the same time?

You have three equations, one involving x and λ, one involving y and λ, and the constraint.