$a,b\in\Bbb R$ $a\cdot b=6$ $\;\min(2a+3b)=?$ 
$a,b\in \Bbb R$
$a\cdot b=6$
$\min(2a+3b)=?$

My Solution:
Substituted $b=\frac{6}{a}$ into $2a+3b$ and took the derivative to determine the minimum point, I got:
$$2-\frac{18}{a^2}=0$$
$$(a_1,b_1)=(-3,-2) \ \ , (a_2,b_2)=(3,2) $$
When these values are plugged in, I get $-12$ as the minimum value of $2a+3b$. But, when I plug in $(-1,-6)$ for $(a,b)$, I get even lower value of $-20$.
The answer is given as $-12$, what am I doing wrong?
 A: The minimum does not exist. 
Try $a\rightarrow-\infty$ and $b=\frac{6}{a}.$
If $a$ and $b$ are integers, so $ab=6$ gives 
$$(a,b)\in\{(1,6),(2,3),(3,2),(6,1),(-1,-6),(-2,-3),(-3,-2),(-6,-1)\}$$
and you can choose the answer.
By the way, if $a$ and $b$ are positives then by AM-GM we obtain:
$$2a+3b\geq2\sqrt{2a\cdot3b}=2\sqrt{6ab}=12.$$
A: Graphical method. You can draw the constraint curves ($b=\frac 6a$) and contour lines ($b=-\frac 23a+\frac c3$):

As contour line shifts lower (blue, red, green, violet lines), it still intersects the constraint black curve (i.e. the constraint is satisfied), hence there is no minimum of $c=2a+3b$. 
But if $a,b>0$, then the minimum of $c=2a+3b$ is achieved at the point $A(3,2)$, that is $c(3,2)=2\cdot 3+3\cdot 2=12$ is minimum. 
A: My solution:
Let $c:=2a+3b$ and eliminate $b$. We have
$$c=2a+\frac{18}a$$ and
$$\frac{dc}{da}=2-\frac{18}a^2=0$$ or $$a=\pm3,$$ which determines the stationary points $c=\pm12$.
$c=12$ is a local minimum (and $c=-12$ a local maximum), but as the function is unbounded near $a=0$, there is no global minimum.
