Bruce's Material Company hauls gravel to a construction site, using a small truck and a large truck. The carrying capacity and operating cost per loadare given in the accompanying table. Bruce must deliver a minimum of 630 cubic yards per day to satisfy her contract with the builder. The unioncontract with her drivers requires that the total number of loads per day is a minimum of 10. How many loads should be made in each truck per day tominimize the total cost?Capacity (yd³)Cost per LoadSmall Truck Large Truck7090$63$76In order to minimize the total cost, the number of loads in a small truck that should be made isbe made is.and the number of loads in a large truck that should

Answered: Bruce's Material Company hauls gravel…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Bruce's Material Company hauls gravel to a construction site, using a small truck and a large truck. The carrying capacity and operating cost per load
are given in the accompanying table. Bruce must deliver a minimum of 630 cubic yards per day to satisfy her contract with the builder. The union
contract with her drivers requires that the total number of loads per day is a minimum of 10. How many loads should be made in each truck per day to
minimize the total cost?
Capacity (yd³)
Cost per Load
Small Truck Large Truck
70
90
$63
$76
In order to minimize the total cost, the number of loads in a small truck that should be made is
be made is.
and the number of loads in a large truck that should

Expert Solution

Step 1

Let number of loads of Small truck per day be X1 and number of loads of Large truck per day be X2.

Min Z

subject to

	70	x1	+	90	x2	≥	630
		x1	+		x2	≥	10

and x1,x2≥0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≥' we should subtract surplus variable S1 and add artificial variable A1

2. As the constraint-2 is of type '≥' we should subtract surplus variable S2 and add artificial variable A2

After introducing surplus,artificial variables

Min Z

subject to

630

and x1,x2,S1,S2,A1,A2≥0

Iteration-1		Cj	76	63	0	0	M	M
B	CB	XB	x1	x2	S1	S2	A1	A2	MinRatio XBx2
A1	M	630	70	(90)	-1	0	1	0	63090=7→
A2	M	10	1	1	0	-1	0	1	101=10
Z=640M		Zj	71M	91M	-M	-M	M	M
		Zj-Cj	71M-76	91M-63↑	-M	-M	0	0

Positive maximum Zj-Cj is 91M-63 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 7 and its row index is 1. So, the leaving basis variable is A1.

∴ The pivot element is 90.

Entering =x2, Departing =A1, Key Element =90

R1(new)=R1(old) ÷90

R2(new)=R2(old) - R1(new)

Iteration-2		Cj	76	63	0	0	M
B	CB	XB	x1	x2	S1	S2	A2	MinRatio XBx1
x2	63	7	(0.7778)	1	-0.0111	0	0	70.7778=9→
A2	M	3	0.2222	0	0.0111	-1	1	30.2222=13.5
Z=3M+441		Zj	0.2222M+49	63	0.0111M-0.7	-M	M
		Zj-Cj	0.2222M-27↑	0	0.0111M-0.7	-M	0

Positive maximum Zj-Cj is 0.2222M-27 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 9 and its row index is 1. So, the leaving basis variable is x2.

∴ The pivot element is 0.7778.

Entering =x1, Departing =x2, Key Element =0.7778

R1(new)=R1(old) ÷0.7778

R2(new)=R2(old) - 0.2222R1(new)

Iteration-3		Cj	76	63	0	0	M
B	CB	XB	x1	x2	S1	S2	A2	MinRatio XBS1
x1	76	9	1	1.2857	-0.0143	0	0	---
A2	M	1	0	-0.2857	(0.0143)	-1	1	10.0143=70→
Z=M+684		Zj	76	-0.2857M+97.7143	0.0143M-1.0857	-M	M
		Zj-Cj	0	-0.2857M+34.7143	0.0143M-1.0857↑	-M	0

Positive maximum Zj-Cj is 0.0143M-1.0857 and its column index is 3. So, the entering variable is S1.

Minimum ratio is 70 and its row index is 2. So, the leaving basis variable is A2.

∴ The pivot element is 0.0143.