October 21, 2009

Perimeter Greater than Area?

Why does our present title then not make any sense? Firstly we are comparing unequal things, like apples and

oranges, or like comparing distance and speed. Obviously you cannot say anything if you are asked which is

greater: distance 100 km or the speed value 200 km per hour?

But nevertheless sometimes interesting information can be gained by pursuing useless things. We are also

doing so to show how elementary questions can be asked and explored at the elementary school level (well, it

helps if children should know graphs).

Okay here is the problem: when is perimeter greater than area for a rectangle (PM being measured say in cm

units and area in sq.cm units)? When we say greater we mean numerically greater.

[Well, if you want to make “sense” of this kind of question, you may reformulate the problem thus: we have

two number crunching machines, the perimeter machine and the area machine. You feed positive numbers a

and b – into both – then at what point the output from the one will exceed the other? ]

Let us take s and t be the sides of the rectangle and PM as its perimeter and A as its area.

1) When s = 1: PM = 2 (length + breadth) = 2 (1 + t). Area is length x breadth = 1x t = t. Now 2 + 2t is

always greater than t. Therefore here PM > A, when s =1.

This means whenever you have one of the sides as 1, perimeter will be always (numerically) greater

than area.

2) When s = 2: again here PM > area as always 2 (2+t) = 4 + 2t > 2t. This means whenever you have s

=2, perimeter will be always (numerically) greater than area, irrespective of whatever value t takes.

3) When s = 3: Here it is bit more curious. PM > A till the other side t is less than 6, and when t > 6, A >

PM, that is area is greater than perimeter. When t is exactly equal to 6, that is when t =6, area and

perimeter are equal (numerically only, mind you). Here A = PM = 18. This is the first instance when

area is equal to perimeter for integer values of the sides (s, t = 3,6). For a given value of s, we will call

the t-value equipoint when area becomes equal to perimeter. Or the point when the relation PM > A,

becomes the other way around, that is A >PM. Again numerically please let us remember.

When s = 4: Till t reaches the value 4, PM > Area. At t = 4, PM = Area. When t > 4, A > PM. When t = 4,

Area = PM =16. Now this is the second instance when area is equal to perimeter for integer values of the sides

(s, t = 4,4). As far as we know (s,t) = (3,6) or (6,3) and (4,4) are the only instances when A = PM for integer

values of the sides. Note the equipoint is t = 4 for s =4. The other integer solutions are,

s= 1, t = -2; s =0, t =0; s = -2, t = 1.

4) When s =5: the equipoint is t = 10/3 =3.333…, that is area = perimeter = 15 in this case. When t

>10/3, A > PM. When t < 10/3, Area < PM.

If you wondered, how we got 10/3, all we did was solve the equation when PM = Area, or: 2 (5 +t) =

5t which leads to 3t =10 or t =10/3.

5) When s = 6: we know from (3) above there is an integer equipoint at t = 3. And also therefore when t

<3, A < PM and when t > 3, A >PM.

6) When s =7: equipoint is t = 14/5 =2.80.

7) When s = 8: equipoint is t = 2.66.

8) When s =9: equipoint is t = 2.57.

9) When s= 10: equipoint is t = 2.50.

You can see the trend: as s increases, the equipoint t (or the point when the relation PM > A, becomes the

other way around, that is A >PM) decreases. Does it indefinitely decrease? Or approaches a limit beyond

which it does not go?

No t does not decrease beyond a point. It approaches a limit: when s increases, the equipoint t decreases to

(or approaches) 2.

How do we know this? We write when PM =A,

2 (s+t) =st

or 2s + 2t = st; or 2s = st –2t = t(s-2). From which we can write t = 2s/(s-2) or

t = 2/(1-2/s). From this it is clear when s increases towards infinity, 2/s tends to zero, or 2/s � 0.

Therefore, t � 2.

We show this in the graphs below, t is on the vertical axis and s is on the horizontal axis. We have plotted

t = 2s/(s-2). In the graphs (produced by Wolfram Alpha), x is our s.

The graph clearly shows as s increases towards positive infinity, it approaches 2 and as s decreases and

goes towards 2 (say from 2.5) then the equipoint t approaches infinity. The latter can be understood when

we calculate t for s = 3, 2.5, 2.2, 2.1, 2.001, 2.0001, etc. The equipoint t values are respectively 6, 10, 22,

42, 4002, 40,002, etc.

If you are further interested in this kind of ‘exciting’ stuff, as s approaches 2 from zero to 1.5, 1.75, 1.90,

the corresponding equipoint, t, approaches minus infinity, etc. Also when s takes values between 0 and

less than 2, the corresponding equipoints t are negative. But when s is negative, that is less than zero; the

corresponding equipoints are always negative.

You may also like to interpret by using your imagination like Alice in Wonderland what do the following

mean or is it pure nonsense:

For example, s = ½, t =-1, A = -½, PM = -1; or s = 3, t =-1, A = -3, PM = 2; or s= -4, t =-5, A = 20, PM = -

18.

That is, area and perimeter both negative with one parameter positive; area negative and perimeter positive

with one parameter positive; or area positive and perimeter negative when both parameters t and s are

negative.

This author cannot think of any meaningful explanation physically. But welcomes suggestions.