10x + 15y <= 70
x >= 3
y >= 0
x is the number of squid ball sticks.
y is the number of cups of noodles.
using the desmos.com calculator, you graph the opposite of these inequalities.
the feasible region is any point pair in the unshaded area of the graph plus any point pair on either of the lines shown that borders the unshaded area of the graph.
only integral points pairs were considered, i.e. neither x nor y could not be an integer.
the graph looks like this:
the point pairs that are in the feasible region aree:
(3,2), (4,2)
(3,1), (4,1), (5,1)
(3,0), (4,0), (5,0), (6,0), (7,0)
take your pick.
any 3 of these will satisfy the requirements of the problem.
the requirements of the problem are that:
10x + 15y <= 70
x >= 3
all points have x >= 3, so the x >= 3 part of the requirements are satisfied.
all points have 10x + 15y <= 70
for simplicity, take the center row.
(3,1) = 3 * 10 + 1 * 15 = 45 which is less than 70.
(4,1) = 4 * 10 + 1 * 15 = 55 which is less than 70.
(5,1) = 5 * 10 + 1 * 15 = 65 which is less than 70.
there was no requirement that she has to spend as much of her money as possible.
the only requirements was that she had to buy at least 3 squid ball sticks and that the total money spent had to be less than or equal to 70.
if you wanted only the point pairs that maximized her expenditures, then you would have to evaluate each of the point pairs for total cost and pick the ones that maximized the cost.
that evaluation would yield the following:
(3,2) = 3 * 10 + 2 * 15 = 30 + 30 = 60
(4,2) = 4 * 10 + 2 * 15 = 40 + 30 = 70
(3,1) = 3 * 10 + 1 * 15 = 30 + 15 = 45
(4,1) = 4 * 10 + 1 * 15 = 40 + 15 = 55
(5,1) = 5 * 10 + 1 * 15 = 50 + 15 = 65
(3,0) = 3 * 10 + 0 * 15 = 30 + 0 = 30
(4,0) = 4 * 10 + 0 * 15 = 40 + 0 = 40
(5,0) = 5 * 10 + 0 * 15 = 50 + 0 = 50
(6,0) = 6 * 10 + 0 * 15 = 60 + 0 = 60
(7,0) = 7 * 10 + 0 * 15 = 70 + 0 = 70
the top 3 points pairs were (4,2), (7,0), (5,1).
those were the ones that maximized her expenditures.
looking at the graph, those were the point pairs that were closest to the line 10x + 15y = 70
the inequalities that were graphed were:
10x + 15y >= 70
x <= 3
y <= 0
these were the opposites of the required inequalities, that were:
10x + 15y <= 70
x >= 3
Y >= 0
i believe i graphed x <= 0 as well, but that was redundant because, if x <= 3 was graphed, than that included x <= 0.