"large and small vans."
L=# of large, L≥0
S=# of small, S≥0
"He can spend no more than $100,000 for both type of vans"
"Kris can purchase a small van for $15,000 and maintain it for $100 per month...a large van for $25,000 and maintain it for $70 per month"
"no more than $600 per month for maintenance"
"large van carries 14 passengers, and each small van carries 7 passengers"
So the above are the constraints and the objective function.
The formulation of the constraints and objective functions is a good exercise, especially when the parameters can change with time.
However, there are times that you don't need a screwdriver to do repairs. Same with linear programming:
Capacity of large van = 2*capacity of small
Cost of large van < 2*cost of small
cost of maintaining large van < 2*cost of maintaining small van
Number of large vans he can buy with budget=100000/25000 = exactly 4 (for 56 passengers)
cost of monthly maintenance = 4*70=280 < 600
So what would be your choice even without the screwdriver?
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