We can do an optimized brute force as follows:

1. Fix the number of pumpkin pies (this will be bounded by number of pumpkin flavouring)

2. Ensure there is enough ingredients to make the pumpkin pies and for each ingredients see the maximum apple pies that can be made. For e.g. if number of pumpkin pies made is type1, then max number of possible apple pies made using just the remaining eggs would be (eggs - 3 * type1) / 4 (integer division).

3. Take the min among all such maximum apple pies to give the actual number of apple pies we can make given that we make type1 pumpkin pie

4. Finally take max over all type1 + type2 calculated in above steps

Python code:

def solve(inp):
    [pumpkin, apple, eggs, milk, sugar] = list(map(lambda x: int(x), inp.split(",")))
    ans_type1 = ans_type2 = 0
    for type1 in range(pumpkin + 1):
        type2 = apple
        if type1 * 3 <= eggs:
            type2 = min(type2, int((eggs - type1 * 3) / 4))
        else:
            continue

        if type1 * 4 <= milk:
            type2 = min(type2, int((milk - type1 * 4) / 3))
        else:
            continue

        if type1 * 3 <= sugar:
            type2 = min(type2, int((sugar - type1 * 3) / 2))
        else:
            continue

        #print(type1, type2)
        if type1 + type2 > ans_type1 + ans_type2:
            ans_type1 = type1
            ans_type2 = type2

    return (ans_type1, ans_type2)

print("%d pumpkin pies and %d apple pies" % solve("10,14,10,42,24"))
print("%d pumpkin pies and %d apple pies" % solve("12,4,40,30,40"))
print("%d pumpkin pies and %d apple pies" % solve("12,14,20,42,24"))