Use Lagrange multipliers to maximize function
$$f(x,y)=6xy,$$
subject to the constraint
$$2x+3y=24.$$
$$F(x,y,\lambda)=6xy+\lambda(2x+3y-24)$$
$$F_{x}=6y+2\lambda=0$$
$$F_{y}=6x+3\lambda=0$$
$$y = - \frac{1}{3} \lambda$$
$$x = - \frac{1}{2} \lambda$$
$$\frac{3}{2}y= - \frac{1}{3} \left( \frac{3}{2} \right) \lambda = - \frac{1}{2} \lambda=x$$
I'm not sure where the $(3y)/2$ came from. Can someone help me with this? I know what to do after. Thank you so much.
You found $x=-\lambda/2$ and $y=-\lambda/3$. Now plug this into the constraint and obtain $$24=2x+3y=-\lambda-3\lambda\ .$$ It follows that necessarily $\lambda=-6$, so that you obtain $x=3$, $\>y=2$. So there is a unique conditionally stationary point $P=(3,2)$.
In order to show that $f(P)=36$ is indeed the global maximum of $f$ for the given constraint some qualitative argument is needed. In this regard note that $f$ is negative in the second and fourth quadrants, and that there are no points fulfilling the constraint in the third quadrant. Therefore it's all about the segment connecting the points $(0,8)$ and $(12,0)$. There is a unique hyperbola $6xy={\rm const.}$ touching this segment, and the corresponding constant $(=36)$ realizes the $\max$ of $f$ under the constraint.