@rob when maximizing a curve $f(x,y)$ by a constraint $g(x,y)=k$ they must both lie on the same tangent plane, right? Doesn't that mean either can be a scalar multiple of the other? So in a sense $\lambda \nabla f(x,y) = \nabla g(x,y) \equiv \nabla f(x,y) = \lambda \nabla g(x,y)$?
@SirCumference I hadn't done so yet ... and after the spate of \renewcommand{\foo}{I'm a doofus} that went on in the sidebar the other day I think I'll refrain.
@rob it wasn't a real question though. I was just wrapping my head around the two functions having to lie on the same tangent plane. only way those two would be equivalent would be $\lambda_2 = \frac{1}{\lambda_1}$
@Obliv I think I've parsed your statement, but I don't understand it straightaway. Why would your function to maximize and its constraint need to share a tangent plane?
like, imagine if g(x,y) = k passed the point marked on the picture then never touched f(x,y) = 11 or f(x,y) = 9 again. Would they still share a tangent plane at f(x,y) = 10?
