Wednesday, August 28, 2013

Choosing a career



He was in a dilemma – one of my close friends – to make a decision for his career! I suppose almost everyone in their late 20s or early 30s come to a point of mid-life career choices and changes.
But this friend was a little different. Instead of letting impulse take over, he applied an analytical process in choosing his career. I learnt about his methods and was personally fascinated by it! Here is his method.
Unlike a fresher searching for a job, it is interesting to see how different a mid-career dynamics work. My friend was no more in a position of distress to search for a job; rather he had a luxury to be offered more jobs than he wanted. That’s where his dilemma started – an expanded list of choices, all seeming good, leaving him pretty confused.
He had four different job offers – all were good in their own aspect. But my friend doesn’t want to settle down for less than the best – he was of course in a critical juncture of his career and personal life. Being a meticulous person, he decided to list out the criteria that he was currently looking for in a job.  Here is his list.
·         Salary: He wanted to make sure that the ‘salary’ was on the list, unlike before. For his previous choices, he knew that ‘salary’ wasn’t a criterion, but now quoting him, ‘also money’ – salary become an important criterion.
·         Location: Neither was location. He has roamed around the globe and this time he decided to stay close to his family.
·         Security: In terms of the job’s nature and continuity for another 10 years at least.
·         Stability: Of the institution
·         Growth: Of the sector, institution and his own skill sets
·         Mentors: Within the institution, within the sector for whom he taught he could seek guidance
·         Match/ Suiting: To his own temperament – whether we like it or not, not all of us are made the same
·         Marriage: Of course he is going to marry soon and doesn’t want his job nature to affect his personal life and vice versa.
As you could see the list was very personal for him and he was careful in listing it and defining it for himself. It could be different for somebody else.
Given the above criteria for the job, he rated each of the job profile he was offered on a scale of 1-3: 1 being bad, 2 being okay and 3 being good.
Here are his ratings:
Criteria
PK
MRT
AFC
TRS*
Salary
2
1
1
3
Location
3
1
2
3
Security
3
3
2
2
Stability
3
1
2
2
Growth
3
2
2
2
Mentors
1
2
3
1
Suiting
2
1
3
1
Marriage
2
1
1
3
Total
19
12
16
17
*Names of the organizations are changed.
Finally, he calculated the total of the ratings. Based on the above analysis and introspection, he did choose the job that rated the highest.
Now it has been a month since he joined the new organization.

Friday, August 9, 2013

The Art of Mathematics



I found this interesting question in one of the articles I was reading on Calculus,
“A rectangle is to be formed by using a piece of wire that is 36 inches long. What will its dimensions be if it encloses the greatest area?”
Here, we have a problem that states, the perimeter of a rectangle is 36 inches, and we have to deduce the dimensions of that rectangle (length and breadth) which will have the greatest area possible.
I have been trying to calculate that.
To cut the long story short, I have made the following table, given the fact that everybody knows that the formula.
Perimeter of a rectangle = 2 (l + b) units
Area of a rectangle = l X b sq. units
With the above conditions, the possible dimensions of the length and breadth should have a sum of l + b = 18
If, 2 (l + b) = 36,
Then, l + b = 18
So the possible dimensions are, 17 + 1, 16 + 2, and 15 + 3… so on

Table: Perimeter v/s Area of the rectangle
P=2(l+b)
36
36
36
36
36
36
36
36
36
l+b
18
18
18
18
18
18
18
18
18
A=lXb
17
32
45
56
65
72
77
80
81
l
17
16
15
14
13
12
11
10
9
b
1
2
3
4
5
6
7
8
9

I have even plotted a graph for that Perimeter against Area. 


With Perimeter being constant at 36 inches, the greatest area is achieved at 81 sq. inches, and the dimensions of length and breadth is 9 inches.
Here is the interesting part, it makes a square. Yes, the rectangle with the greatest area is a ‘square’.
I tried working out for other possible values of perimeter as well. For every problem, it comes that “area with the greatest rectangle is a square, for the given value of perimeter.”
So what?
If you  have constructing or living in a home, we could now deduce that best possible room with the maximum amount of space/ area will be a square with the same amount of mortar used (perimeter value).
But we don’t usually build square rooms, do we?
On the other hand, it has been a very common practice among us, as told by our forefathers, the best possible dimensions of a room will be a product of two consecutive odd numbers, say – 9 X 11, 11 X 13… etc… This has serious amount of scientific reasoning into it, for the product of two consecutive odd numbers is the ‘second best’ spacious dimension for the given amount of perimeter.
Interesting isn’t it?
So two conclusions that has kept me thinking were,
  1. The possibility that there could be two rectangles with same perimeter but varying area
  2. A square is the largest rectangle with maximum area for the given perimeter 
Oh the beauty of Maths!!