Why does a square building have less cost per square meter than a rectangular one of the same area?

Answer 1

At a guess i'd say it was because a rectangular building would would have a larger perimter in all cases, so if it cost more to build the outward-facing walls of a building, then having a square building would reduce this cost.

To prove a square always has the least perimeter per area:

Consider a rectangle with length l and width w. We can write the following about it's area, A and perimeter, P:

(1) A = lw

(2) P = 2l + 2w.

To prove that the rectangle has sides all of equal length (w = l), I will find the minimum perimeter, P (and the corresponding ratio of l and w) for a fixed area, A.

From equations (1) and (2),

(3) w=A/l

(4) P = 2l + 2A/l.

For very large and small values of l, the perimeter is bigger than the optimum (for example, a rectangle has area 9, if l is very small/large, (0.9/10), then the perimeter is large (21.8) compared to a moderate value of l (l = 3, perimeter is 12) [calculated using equation (4)]. So, if P is large for very big and very small values of l, and gets smaller in the middle (quite like a U shape), there must be a point where P stops getting smaller and gets bigger again (the tip of the U). This is called the 'turning point' of P.

The quantity 'dP/dL' is the rate of change of P as l changes. This is equal to 0 at the turning point of l. An algabraic expression for dP/dl can be found by following a few simple rules (look for 'differentiation by first principles' to have these explained):

(5) dP/dl = 2 - 2A/l²

So our minimum value of P is at the turning point of P, when dP/dl = 0, solving this gives:

dP/dl = 0

=> 2 - 2A/l² = 0

=> A/l² = 1

=> A = l² (6)

Recall from equation (3) that the width of the rectangle, w is given by

(3) w = A/l

So

w = A/l = l²/l = l.

So the width is the same as the length of the rectangle when the perimeter is at a minimum, thus the rectangle must be a square.