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August 20, 2014

August 20, 2014

Posted by **fakaapo** on Thursday, April 14, 2011 at 8:09am.

Kris is trying to make his business more efficient by having a system of both large and small vans. He can spend no more than $100,000 for both type of vans and no more than $600 per month for maintenance. Kris can purchase a small van for $15,000 and maintain it for $100 per month. He can purchase a large van for $25,000 and maintain it for $70 per month. Each large van carries 14 passengers, and each small van carries 7 passengers. Kris is interested in knowing how many of each van he should purchase so that he can serve the maximum number of passengers.

- math -
**MathMate**, Thursday, April 14, 2011 at 10:18am"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"

25000L+15000S≤100000

"no more than $600 per month for maintenance"

70L+100S≤600

"large van carries 14 passengers, and each small van carries 7 passengers"

Utility=objective function=

P(S,L)=7S+14L

So the above are the constraints and the objective function.

Note:

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:

Here:

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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