Tom K. answered 01/25/21
Knowledgeable and Friendly Math and Statistics Tutor
Let the large float be L and the small float be S
There are 10 ft behind each float except for the last one.
a) 30L + 10L + 15S + 10S - 10 = 40L + 25S - 10
b) Label points (L, S)
We can think of a lot of shortcuts, but we see that solutions will have L ranging from 0 to 5 and then finding what values of S solve the equation, as 40L - 10 <= 200 means 40L <= 210, or L <= 5
L = 0. 150 <= 25S - 10 <= 200 160 <= 25S <= 210 S = 7, 8
L = 1. 150 <= 25S + 40 - 10 <= 200 120 <= 25S <= 170 S = 5, 6
L = 2. 150 <= 25S + 80 - 10 <= 200 80 <= 25S <= 130 S = 4, 5
L = 3. 150 <= 25S + 120 - 10 <= 200 40 <= 25S <= 90 S = 2, 3
L = 4. 150 <= 25S + 160 - 10 <= 200 0 <= 25S <= 50 S = 0, 1, 2
L = 5. 150 <= 25S + 200 - 10 <= 200 -40 <= 25S <= 10 S = 0
Solutions: (0, 7), (0, 8), (1, 5), (1, 6), (2, 4), (2, 5), (3, 2), (3, 3), (4, 0), (4, 1), (4, 2), (5, 0)
The budget is $2500. Floats are $300 for small and $600 for the large. Thus, append this to the values above.
(0, 7, 2100), (0, 8, 2400), (1, 5, 2100), (1, 6, 2400), (2, 4, 2400), (2, 5, 2700), (3, 2, 2400), (3, 3, 2700), (4, 0, 2400), (4, 1, 2700), (4, 2, 3000), (5, 0, 3000)
Feasible solutions have cost less than or equal to $2500
(0, 7, 2100), (0, 8, 2400), (1, 5, 2100), (1, 6, 2400), (2, 4, 2400), (3, 2, 2400), (4, 0, 2400)