Lagrange Multipliers: Find $\min$ of $f(x,y)=3(x+1) +2(y-1)$ subject to the constraint $x^2+y^2=4$ 
Find the minumum value of the function $f(x,y)=3(x+1) +2(y-1)$, subject to the constraint that $x^2+y^2=4$.

The problem states to use Lagrange Multipliers. In doing so I obtained the point $(\frac 6{\sqrt{13}},\frac 4{\sqrt{13}})$. But when checking if it is a maximum I found that the result was inconclusive. What did I do wrong?

Write: $f(x,y) = 3x+2y + 1$. For a quick answer, using Cauchy-Schwarz inequality: 
$|3x+2y| \leq \sqrt{3^2+2^2}\sqrt{x^2+y^2} = 2\sqrt{13}$. Thus:
$-2\sqrt{13}\leq 3x+2y \leq 2\sqrt{13} \to 1-2\sqrt{13} \leq f(x,y) \leq 1+2\sqrt{13}$.
