# minimization problem

• Dec 12th 2005, 06:14 PM
cobragrrll
minimization problem
You must deliever a min. of 200 cubic yards per day to satisfy a union contract. The union contract requires that the total number of loads per day is a min. of 6. How many loads should be made in each truck per day to minimize the total cost?

Small Truck Large Truck
Capacity (yd^3) 40 60

• Dec 13th 2005, 01:28 AM
CaptainBlack
Quote:

Originally Posted by cobragrrll
You must deliever a min. of 200 cubic yards per day to satisfy a union contract. The union contract requires that the total number of loads per day is a min. of 6. How many loads should be made in each truck per day to minimize the total cost?

Small Truck Large Truck
Capacity (yd^3) 40 60

Since 6 trucks a day of either kind will deliver more than the
required minimum 200 cubic yds (240 and 360 cubic yds respectively
for the small and large trucks). The cost is minimised by using
6 truck of whichever size has the least cost per load.

In this case it is 6 large trucks, which deliver 360 cubic yards
per day for a cost of \$258.

RonL
• Dec 13th 2005, 04:16 AM
ticbol
Quote:

Originally Posted by cobragrrll
You must deliever a min. of 200 cubic yards per day to satisfy a union contract. The union contract requires that the total number of loads per day is a min. of 6. How many loads should be made in each truck per day to minimize the total cost?

Small Truck Large Truck
Capacity (yd^3) 40 60

Umm, another linear programming, as I see it.

Decision variables:
s = number of small truck to use per day
L = number of Large truck to use per day.

Problem constraints:

Per minimum required volume per day,
40s +60L >= 200
2s +3L >= 10 ------------(1)

Per minimum required number of number of loads per day,
s +L >= 6 -----------(2)
The problem did not specify that both small and large trucks must be used. Just minimum of 6 loads---using whatever trucks, apparently.

Non-negative constraints,
s >= 0 ------------------------(3)
L >= 0 ---------------(4)

Objective function:
To minimize cost.
Total Cost, C = s*\$61 +L*\$43
C = 61s +43L -------in dollars.

Using rectangular s,L axes, where horizontal axis is for values of s, and vertical axis is for values of L, graph the 4 linear inequalities above. Find the corner points of the feasible region.

You should arrive at:
>>>feasible region is not fully bounded. Its only boundaries are the s and L axes, and the linear inequality (2), and then it is open up to before positive infinity.
Although inqualities (1) and (2) intersect at (8,-2), this point is not a corner point of the feasible region because consntraint/inequality (4) says L >=0.
>>>Corner points are (0,6) and (6,0) only.

Test objective function on those corner points,
---at (0,6), C = 61*0 +43*6 = \$258.
---at (6,0), C = 61*6 +43*0 = \$366

The (0,6) gave the lower C, therefore, to minmize total cost per day, deliver 6 loads by using only the large trucks. -------answer.