Raymond B. answered 01/28/21
Math, microeconomics or criminal justice
4F + 6S < 30, divide by 2
2F + 3S < 15
F+S > 4
2F + 2S > 8
8< 2F+2S < 2F + 3S < 15
Graph 2 lines, using the above equations, ignoring the inequalities, and you get 2 downward sloping lines.
One goes from (0,4) to (4,0) the other lies above that going from (0,7.5) to (5,0)
the area between those two lines to the right of the y axis and above the x axis that includes the solutions to the constraints
Since the solution involves only integers. it's only the integers within that area that are feasible
5F, 0S
4F, 0 or 1S
3F, 1, 2 or 3S
2F, 2 or 3S
1F, 3,4,5 or 6S
0F, 4,5, or 6 S
15 possibilities within the budget and at least 4