# Optimisation using constraints

• lagwagon555
In summary, we are using Lagrange multipliers to find the highest and lowest points on the curve of intersection between an elliptic paraboloid and a right circular cylinder. The Lagrangian is set up and equations for Lx and Ly are derived. However, we encounter a problem with the equation involving λ and need to explore both possibilities for y. The answer of x = 1/3, y = +-0.94 is obtained, but there is doubt about its correctness and the method is suggested to be tried directly as well. It is noted that the y=0 case must also be considered.

## Homework Statement

Consider the intersection of two surfaces: an elliptic paraboloid
z = x2 + 2x + 4y2 and a right circular cylinder x2 + y2 = 1. Use Lagrange multipliers to find
the highest and lowest points on the curve of intersection

## The Attempt at a Solution

L = x^2 + 2x + 4y^2 - λ(x^2 + y^2 - 1)
Lx = 2x + 2 - 2λx = 0
Ly = 8y - 2λy = 0

Rearranging gives the bizarre result of λ = 4. Aren't I supposed to be able to eliminate the lagrange multipliers by it giving me λ in terms of y? It's giving me a value instead.

I ran with this anyway, and got the answer of x = 1/3, y = +-0.94, but I don't think this is correct. I suspect I'm doing something wrong algebraically? Any help hugely appreciated :)

lagwagon555 said:
I suspect I'm doing something wrong algebraically? Any help hugely appreciated :)

You need to be careful with the last equation. I believe you divided by y, but that's valid only if y isn't zero. You need to explore both possibilities. By the way, you forgot an equation. What about Lλ?

What makes you think you have the wrong answer?

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If you don't trust the method try it directly as well. Use the constraint to eliminate y^2 in z. Now look at the extrema for x in [-1,1]. You do need to consider the y=0 case as well.