Question #04991

Well, a leaf has relatively little thickness. So, we can approximate its SA/V ratio to be #1:1# (which becomes more apparent if the leaf is flattened). More realistically it is probably slightly larger, maybe #1.01:1#-ish.

On the other hand, a skyscraper can be approximated as a rectangular prism, with volume

#V = L xx W xx H#

and surface area

#SA = 2(L xx W) + 2(L xx H) + 2(W xx H)#

where #L#, #W#, and #H# are length, width, and height.

For this approximation then:

#SA"/"V ("skyscraper") ~~ (2LW + 2LH + 2WH)/(LWH)#

To prove this claim that the SA/V ratio for a leaf is more than that of a skyscraper, we need to show that this ratio is less than #1#.

In general, the length and width are usually similar in comparison to the height, which is much larger for a tall skyscraper, so we assume #L ~~ W# for simplicity to get:

#SA"/"V ("skyscraper") ~~ (2L^2 + 2LH + 2LH)/(L^2H)#

#= (2L^2 + 4LH)/(L^2H)#

#= ((2L^2)/H + 4L)/(L^2)#

#= (2)/H + 4/L#

When a skyscraper is particularly tall, #2/H = 1/H + 1/H ~~ 0#, so:

#color(blue)(SA"/"V ("skyscraper") ~~ 4/L < 1/1)#

So, the surface area to volume ratio of the average skyscraper is less than #1#, the ratio for the average leaf.

Let's say the average skyscraper is around #"100 m"# tall with a square base of length about #"20 m"#. Then,

#SA"/"V ~~ (2(20)(20) + 2(20)(100) + 2(20)(100))/(20cdot20cdot100) "m"^(-1)#

#= "0.22 m"^(-1) < 1#

And our final approximation would give:

#SA"/"V ~~ 4/L = 4/20 "m"^(-1) = "0.20 m"^(-1) < 1#