This is called "linear programming".
let i = number of Irish
let b = number of banana
then profit = p = 1.5 i + 4 b
now lets plot i on the x axis and b on the y axis. For every value of p there is a line on that graph of form:
b = (-1.5/4)i + p/4
b = -.375 i + .25 p
NOW, find the feasible region on the graph
You only have 16 flours, so there is a line going from (0,4) (8,0). Call that the flour limit line and draw it on your graph
You only have 3 sugars, so there is a line going from (0,3) to (12 ,0). Call it the sugar limit line and draw it on the graph.
the sugar line hits the flour line where?
flower line b = 4 - .5 i
sugar line b = 3 - .25 i
solve (you could get this from your graph of course)
0 = 1 -.25 i
i = 4
b = 2
NOW, we must test the corners for maximum p
p(0,0) = 0
p(0,3) = 1.5(0)+4*3 = 12
p(4,2) = 1.5*4 + 4*2 = 14
p(8,0) = 1.5(8) +4(0) = 12
max profit = 14 at i = 4 and b = 2
This is a "linear programming" problem.
Let the number of Banana bread be x
and the number of Irish bread be y
from the flour limitation we have
4x + 2y ≤ 16
2x + y ≤ 8
from the sugar limitation we have
(1/4)x + y ≤ 3
x + 4y ≤ 12
when these two are graphed in the first quadrant of a graph, we get a region bounded by the origin, the x and y intercepts closest to the origin and the intersection of the corresponding equations.
The profit equation would be
P = 4x + 1.5y
the slope of that line is -8/3
The farther this line can move away from the origin (a profit of zero) while still within our region, the larger the profit.
So we can move as far as the intersection of
2x+y = 8 and x+4y = 12
I get y = 16/7 but how can we bake 16/7 loafs of bread?
so let y be the closest whole number or y = 2, then x = 3
the profit would be 3(4) + 2(1.5) = 15
since both x and y must be whole numbers, there are only 5 possible cases
(0,8), (1,6), (2,4), (3,2), and (4,0)
It would be easy to see that (3,2) produces the largest profit.