Hello everyone! It's me Shouq. This is the first lab report for an Industrial Engineering class called Operational research / optimization - OR1 - IE335 which I took in Fall 2017 in the American University of the Middle East (AUM) in Kuwait.
Result and discussion
Area Optimization
Abstract
The objective of this report is to show our work in how to find the
maximum area of a rectangle by using an operational research approach. At
first, we will identify the decision variables, and find the objective
functions as well as the constraints. We then try to find the domain of the
objective function and the critical points. Also, we will sketch a graph of the
objective function using both basic calculus knowledge and Excel.
Keywords: Maximum, area,
operational, research, rectangle.
Introduction
Operational resource approach is the way we
are using to find the objective of this report which is to find the maximum
area of a rectangle. Operational resource approach is a scientific way to find
the best requirement and solve problems under certain constraints. Also, Decision
Variables are what will be affecting the final result which will be discussed
in a bit. On the other hand, objective functions and criterions can be either maximum
or minimum.
For
simplicity let us take L = 1 m = 100 cm
Decision Alternative: Trying to decide the width and the height that will give us the
maximum area.
Restrictions and constraints: L = 2 ( H + W )
Width and height cannot
logically be negative. W >= 0, H >= 0
Solution: Let Z be the area of the rectangle. Z = W * H
General OR model:
Max
Z = W * H
Subject
to L = 2 ( H + W )
W >= 0, H >= 0
Where Z is the area, W is the
width of the rectangle, H is the high of the rectangle, and finally L is the
wire length that we assumed to be 1 m = 100 cm in this case.
Result and discussion
In each rectangle we notice that we have two height lines and two width
lines. In the table below we notice that the height here is actually the total
of the two height lines added together and same goes to the width.
Length (L)
|
Height (H)
|
Width
|
Area
|
100
|
99
|
1
|
99
|
98
|
2
|
196
|
|
97
|
3
|
291
|
|
96
|
4
|
384
|
|
95
|
5
|
475
|
|
94
|
6
|
564
|
|
93
|
7
|
651
|
|
92
|
8
|
736
|
|
91
|
9
|
819
|
|
90
|
10
|
900
|
|
89
|
11
|
979
|
|
88
|
12
|
1056
|
|
87
|
13
|
1131
|
|
86
|
14
|
1204
|
|
85
|
15
|
1275
|
|
84
|
16
|
1344
|
|
83
|
17
|
1411
|
|
82
|
18
|
1476
|
|
81
|
19
|
1539
|
|
80
|
20
|
1600
|
|
79
|
21
|
1659
|
|
78
|
22
|
1716
|
|
77
|
23
|
1771
|
|
76
|
24
|
1824
|
|
75
|
25
|
1875
|
|
74
|
26
|
1924
|
|
73
|
27
|
1971
|
|
72
|
28
|
2016
|
|
71
|
29
|
2059
|
|
70
|
30
|
2100
|
|
69
|
31
|
2139
|
|
68
|
32
|
2176
|
|
67
|
33
|
2211
|
|
66
|
34
|
2244
|
|
65
|
35
|
2275
|
|
64
|
36
|
2304
|
|
63
|
37
|
2331
|
|
62
|
38
|
2356
|
|
61
|
39
|
2379
|
|
60
|
40
|
2400
|
|
59
|
41
|
2419
|
|
58
|
42
|
2436
|
|
57
|
43
|
2451
|
|
56
|
44
|
2464
|
|
55
|
45
|
2475
|
|
54
|
46
|
2484
|
|
53
|
47
|
2491
|
|
52
|
48
|
2496
|
|
51
|
49
|
2499
|
|
50
|
50
|
2500
|
|
49
|
51
|
2499
|
|
48
|
52
|
2496
|
|
47
|
53
|
2491
|
|
46
|
54
|
2484
|
|
45
|
55
|
2475
|
|
44
|
56
|
2464
|
|
43
|
57
|
2451
|
|
42
|
58
|
2436
|
|
41
|
59
|
2419
|
|
40
|
60
|
2400
|
|
39
|
61
|
2379
|
|
38
|
62
|
2356
|
|
37
|
63
|
2331
|
|
36
|
64
|
2304
|
|
35
|
65
|
2275
|
|
34
|
66
|
2244
|
|
33
|
67
|
2211
|
|
32
|
68
|
2176
|
|
31
|
69
|
2139
|
|
30
|
70
|
2100
|
|
29
|
71
|
2059
|
|
28
|
72
|
2016
|
|
27
|
73
|
1971
|
|
26
|
74
|
1924
|
|
25
|
75
|
1875
|
|
24
|
76
|
1824
|
|
23
|
77
|
1771
|
|
22
|
78
|
1716
|
|
21
|
79
|
1659
|
|
20
|
80
|
1600
|
|
19
|
81
|
1539
|
|
18
|
82
|
1476
|
|
17
|
83
|
1411
|
|
16
|
84
|
1344
|
|
15
|
85
|
1275
|
|
14
|
86
|
1204
|
|
13
|
87
|
1131
|
|
12
|
88
|
1056
|
|
11
|
89
|
979
|
|
10
|
90
|
900
|
|
9
|
91
|
819
|
|
8
|
92
|
736
|
|
7
|
93
|
651
|
|
6
|
94
|
564
|
|
5
|
95
|
475
|
|
4
|
96
|
384
|
|
3
|
97
|
291
|
|
2
|
98
|
196
|
|
1
|
99
|
99
|
From the graph above, we notice that the maximum area of a rectangle of
100cm length will be 2500. When the area is equal to 2500, the heights added
together will be equal to 50cm ( 25cm + 25cm ) as well as the width added
together will be equal to 50cm ( 25cm + 25cm ).
The problem in one decision
variable:
A
= H * W
L
= 2 * ( H + W)
L
/ 2 = W + H
W
= ( L / 2 ) – H
Area
Z = W * H
= ( ( L / 2 ) – H ) * H
= ( L / 2 ) * H – H^2
If we toke a number for H, we can simply find the area Z as we already
know that L is equal to 100 cm.
To find the maximum Area:
dA
/ dH = 0
dA
/ dH ( ( L /2 )H – H^2 ) = 0
L
/ 2 – 2*H = 0
2*H
= L / 2
H
= L / 4
W
= L / 2 – H
(
L / 2 ) – ( L / 4 ) = ( L / 4 )
(
L / 4 , L / 4 ) is optimal solution
dA
/ dH ( L / 2 – 2*H ) = 0
0 – 2 = 0
No
maximum or minimum of the objective function
Find
critical point:
Max
f(x,y) = x * y
2
* ( x + y ) = L
x
>= 0, y >=0
x
+ y = L/2
y
= L/2 – x x = L/2 – y
From
y = L/2 – x in f(x,y)
F(x,y)
= x * (L/2 – x)
= L/2 * x – x^2
∂ f(x,y) / ∂ x = ( L / 2 ) * x – x^2
For the max of x:
L/2
– 2 * x = 0
2*x
= ( L / 2 ) * 2
4*x
= L
x
= L/4
∂ f(x,y) / ∂ y = ( L / 2 ) – 2 * y
For the max of y:
(L/2
– y ) * y = f(x,y)
f(x,y)
= L/2 * y – y^2
∂ f(x,y) / ∂ y = ( L / 2 ) – 2 * y = 0
y = L / 4
∂ f (w,h) / ∂ w = ( L / 2 ) – 2*W
∂ f (w,h) / ∂ h = ( L / 2 ) – 2*h
In this problem, we assumed that the
length is equal to 1 m which is 100 cm.
The
domain of the objective function:
Z = W * H
= ( ( L / 2 ) – H ) * H
= H * ( L / 2 ) – H^2
H / 2 – H^2 >= 0 H
* ( ( L / 2 ) – H ) >= 0
H^2 =< ( H / 2 )
Sqrt(H^2) =< ( H / 2 )
H =< Sqrt(H/2)
H > 0, H < 1 / ( sqrt(2) )
Z = W * H
= ( ( L / 2 ) ) – H * H
( L / 2 – H ) – H >= 0
H = 0, H = L / 2
[ 0, L/2 ]
Conclusion
In conclusion, we can use the operation research approach in order
to find the general operational research model. It is very helpful if we are
trying to find solution to problems regarding minimum or maximum. Also, the
operational research model will help us to make better decisions and find
optimal solutions.
In this report, we used the operational resource approach to find
the maximum area of a rectangle that has 100 cm length. Moreover, the problems
are broken down into basic components and then solved in defined steps by
mathematical analysis.
The students who worked in this lab report are
- Shouq Alansari
- Hanan Akbar
- Manal Al-Mutairi
- Nour Almuwai
- Reem Almertiji
Knowing that not all students in the group have put equal efforts on this lab report. Some students worked harder than others, and it is normal when it comes to working on groups.
Things to learn from this lab report:
- Be aware of the "number" of spaces you have. Your professor will detect marks if "the number of spaces" after the first paragraph is different than the "number" of spaces after the second paragraph.
- Always write a comment under any graph, table, or picture you have.
- Be professional as much as you can.