Quebec Secondary 5 math tutor, online

The program adds two expectations of any switching student: they "should be able to work independently and feel comfortable with the approach and requirements of the new option", and, under the Secondary V teacher's supervision, "must also be willing to devote all the time they need" to catch up. Our reading, labelled as ours: a summer of focused work on the bridge list above is the honest minimum for CST to TS, and the school's own plan is the starting point for CST to Science.

The problem (ours). A winter carnival committee carves two kinds of ice sculpture. A small one takes 2 hours of carving and 1 block of ice; a large one takes 3 hours and 3 blocks. The committee has 24 hours of volunteer carving and 18 blocks. Judges award 2 visitor points for a small sculpture and 5 for a large one. How many of each should the committee carve to earn the most points?

1. Name the variables. Let s be the number of small sculptures and L the number of large ones, with s at least 0 and L at least 0.
2. Write the constraints. Carving time: 2s + 3L is at most 24. Ice: s + 3L is at most 18. These two inequalities, with the two non-negativity ones, are the system of first-degree inequalities.
3. Find the polygon of constraints. Its corners are (0, 0), (12, 0) where the time line meets the s-axis, (0, 6) where the ice line meets the L-axis, and the crossing of the two lines: subtract s + 3L = 18 from 2s + 3L = 24 to get s = 6, then L = 4.
4. Test the function to be optimized at each corner. Points P = 2s + 5L give 0, 24, 30 and, at (6, 4), 12 + 20 = 32.
5. Decide, and say why. Carve 6 small and 4 large sculptures, for 32 points. It uses all 24 hours (12 + 12) and all 18 blocks (6 + 12), which is why the answer sits on the corner where both constraints are tight.

# Our linear programming check: winter carnival ice sculptures.
# s small sculptures: 2 h of carving, 1 block of ice, 2 visitor points
# L large sculptures: 3 h of carving, 3 blocks of ice, 5 visitor points
HOURS, BLOCKS = 24, 18

def points(s, L):
    return 2 * s + 5 * L

feasible = [(s, L) for s in range(0, 13) for L in range(0, 7)
            if 2 * s + 3 * L <= HOURS and s + 3 * L <= BLOCKS]
top = max(points(s, L) for s, L in feasible)
print("feasible plans:", len(feasible))
print("most points:", top, "at", [p for p in feasible if points(*p) == top])

for corner in [(0, 0), (12, 0), (6, 4), (0, 6)]:      # vertices of the polygon
    print(corner, points(*corner))

feasible plans: 54
most points: 32 at [(6, 4)]
(0, 0) 0
(12, 0) 24
(6, 4) 32
(0, 6) 30

Where students lose marks on problems like this, in our experience: reversing an inequality when writing a constraint, testing only some of the corners, and stating "(6, 4)" without saying in words what it means for the committee. The last one costs most, because the decision is the point of the exercise.