Calculus/Linear Programming Question?

Question

Question: A car dealer has two warehouses, one in Millville and one in Trenton, and two dealerships, one in Camden and one in Doldrum. Every car sold at the dealerships is delivered from the warehouses. On a certain day, Camden sells 10 cars, Doldrum sells 12. Millville warehouse has 15 cars available, and Trenton has 10. The cost of shipping one car is $50 from Millville to Camden, $40 from Millville to Doldrum, $60 from Trenton to Camden and $55 from Trenton to Doldrum. How can the dealer best reduce shipping costs? Namely, how many cars should be moved from each warehouse to each dealership?

I understand there will be 4 variables and I need to reduce to two.

Hint given: the two variables that remain should be the # of cars shipped from the same warehouse (one or the other) to the two dealerships.

The 4 vertices - values - 1080, 1050, 1085, & 1130.

Help!

Thanks!

Shilo

Answer 1

Send A cars from Millville to Camden, B from Millville to Doldrum, C from Trenton to Camden, and D from Trenton to Doldrum.

Then A + C = 10 and B + D = 12. A, B, C, D ≥ 0, so 0 ≤ A ≤ 10 and 0 ≤ B ≤ 12. But also A + B ≤ 15, and C + D ≤ 10 => 10 - A + 12 - B ≤ 10 => A + B ≥ 12.
The objective function to be minimised is
50A + 40B + 60C + 55D
= 50A + 40B + 60(10-A) + 55(12-B)
= 1260 - 10A - 15B.

The corners of the feasible region are (0, 12), (10, 2), (3, 12) and (10, 5), with corresponding objective function values 1080, 1130, 1050 and 1085 respectively. The minimum is 1050 so the optimal distribution is A = 3, B =12, in other words:
3 cars from Millville to Camden
12 cars from Millville to Doldrum
7 cars from Trenton to Camden
0 cars from Trenton to Doldrum.