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?
 A: Let's see. Writing $\nabla f = \lambda \nabla g$, we have
$$3 = \lambda . 2x \ \ \text{ and } \ \ 2 = \lambda . 2y$$
The multiplier $\lambda \neq 0$ and hence $x = 3/(2\lambda)$ and $y = 1/\lambda$. Substituting back into the constraint $g$, $\displaystyle \frac{1}{4\lambda^2} ( 9 + 4) = 4$ and we find $\lambda = \pm \sqrt{13}/4$. In the negative case $\lambda = -\sqrt{13}/4$,
$$f\left(-\frac{6}{\sqrt{13}},-\frac{4}{\sqrt{13}}\right) = 1 - 2\sqrt{13}$$
which is also the global minimum of $f$. Intuitively, this is appealing as $\nabla f = (3,2)$, pointing away from this solution. More formally, we need to analyze the Hessian matrix, as indicated in one of the other answers.
A: 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:
$-2\sqrt{13}\leq 3x+2y \leq 2\sqrt{13} \to 1-2\sqrt{13} \leq f(x,y) \leq 1+2\sqrt{13}$.
A: $$minimize \hspace{3mm} 3(x+1) + 2(y-1)$$
$$s.t. \hspace{3mm} x^2+y^2=4.$$
The lagragian function is $\mathcal{L} = 3(x+1) + 2(y-1) + \lambda (x^2+y^2-4)$. Thus,
$$\frac{\partial\mathcal{L} }{\partial x} = 3 + 2\lambda x = 0 \Rightarrow \lambda = \frac{-3}{2x},$$ and
$$\frac{\partial\mathcal{L} }{\partial y} = 2 + 2\lambda y = 0 \Rightarrow \lambda = \frac{-2}{2y}.$$ Then, $x = \frac{3y}{2}$.
Substituting in the restriction,
$\frac{9y^2}{4}+ y^2 = 4 \Rightarrow y = \pm \frac{4}{\sqrt{13}},$ and $x = \pm \frac{6}{\sqrt{13}}.$ We have two critical points.
For $(x,y) = (\frac{6}{\sqrt{13}},\frac{4}{\sqrt{13}}), \lambda = -\sqrt{13}/4$.
For $(x,y) = (-\frac{6}{\sqrt{13}},-\frac{4}{\sqrt{13}}) , \lambda = \sqrt{13}/4$.
The Hessian is 
\begin{equation}
H =  \left [ \begin{array}{cc}
2\lambda & 0 \\
0 & 2\lambda \\
\end{array} \right]
\end{equation} and
For $\lambda = \sqrt{13}/4$, the Hessian is a positive definite matrix, and for $\lambda = -\sqrt{13}/4$, the Hessian is a negative definite matrix (we don't need analyse the Hessian on the tangent space). Thus, we have a minimal and a maximum point.
By substituting both points in cost function $f(x,y)$ we have,
$f(\frac{6}{\sqrt{13}},\frac{4}{\sqrt{13}}) \approx  8.2111$ (global maximum).
$f(-\frac{6}{\sqrt{13}},-\frac{4}{\sqrt{13}}) \approx  -6.2111$ (global minimum).
