Hello, DooBeeDoo!
Constraint [4] puts us in Quandrant 1.
For , graph the line: .
It has intercepts: (40, 0), (0, 60)
Draw the line and shade the region below it.
For graph the line: .
It has intercepts: (80, 0), (0, 20)
Draw the line and shade the region below it.
The shaded region looks like this . . . Code:

60 *
*
 *
 *
 *
 *
20 o *
:::* *
:::::::o
::::::::* *
:::::::::* *
::::::::::* *
  o      o      *  
 40 80
Constraint [3] is useless.
The line has intercepts (50, 0), (0, 100)
. . and falls outside of the shaded region.
We are concerned with the vertices of the shaded region.
We can see three of them: .(0, 0), (40, 0), (0, 20)
We find the fourth by locating the intersection of the two slanted lines.
We solve the system: . 32,\:12)" alt="\begin{array}{ccc}3x + 2y &=&120 \\ x + 4y &=& 80\end{array}\;\text{ and get: }\32,\:12)" />
We substitute these coordinates into the Zfunction
. . to determine which one maximizes Z.