Find Extrema Subject to Constraint: f(x, y) = x - y, and....

[merlin2007]

Hi everyone,

I'm having a problem with finding extrema for the following function, where the system of equations is singular.
\(\displaystyle \[f\left( {x,y} \right) = x - y\]\), subject to \(\displaystyle \[x^2 - y^2 = 2\]\).

When we take the gradient of both and equalize with the Lagrange multiplier, we get
\(\displaystyle $\begin{array}{l} \nabla h = \left( {1, - 1} \right) - \lambda \left( {2x, - 2y} \right) = {\bf{0}} \\ 1 = 2x\lambda \\ - 1 = - 2y\lambda \\ \end{array}$\)

However, when we then substitute this result into our constraint equation, we get that 0 = 2.
Could someone please provide an interpretation for this result?

Thanks a lot!

 

[tkhunny]

Re: Find Extrema Subject to Constraint

Saddle Point?
Unbounded?
Do you have conditions under which you expect it not to work?

Analytic Geometry Review: What are the equations of the asymptotes of the hyperbola x^2 - y^2 = 2?

 

[soroban]

\(\displaystyle \text{Hello, merlin2007!}\)


\(\displaystyle \text{Note that: }\,f\left( {x,y} \right) = x - y\,\text{ and }\,x^2 - y^2 = 2\,\text{ do not intersect.}\)