Lagrange Multipliers have no solution

Question

$f(x,y)=2x+y$ subject to constraint $x+y=m$. $(2,1)=\lambda(1,1)$ but this does not satisfy $x+y=m$ So there are no solution?

Answer 1

You are correct, there are no solution. It is pretty obvious that $x+y = m$ represent a line in the plane and $2x+y$ is a nontrivial linear function on this line. It is impossible to have critical point.

What's more, just substitute $x+y=m$ into $2x+y$ give $m+x$, and $x$ can be anything........

Answer 2

Another way of saying what is at issue here: the Lagrange-multiplier method is intended to locate points on the curve, surface, etc. where the gradient vector $ \ \nabla f \ $ has the same direction as the gradient for the constraint function, $ \ \nabla g \ $ . (Thus, one is at least a scalar multiple $ \ \lambda \ $ of the other, giving this technique its name.)

But in this problem as written, the normal vector to the line $ \ f(x,y) \ = \ 2x + y \ $ is always in a different direction from the normal vector to the line of the constraint function, $ \ x + y \ = \ m \ . $ So the method will not produce a solution. [You wound up with the "equation" $ \ \langle 2, 1 \rangle \ = \ \lambda \ \langle 1, 1 \rangle \ . \ $ As they say: "not gonna happen"...]