as you increase an object's size, you are increasing its volume by the increase cubed, volume is a 3D quantity, and its surface are by the increase squared surface area is a 2D quantity.
In the same way, if the cell size is decreased, its volume and surface area will also decrease, but at unequal rates. Again, the volume would decrease faster than the surface area, leading to an increase of the surface unit to volume ratio. Hence, as cell size decreases, the surface area to volume ratio increases.
As you increase an object’s size, you are increasing its volume by the increase cubed (volume is a 3D quantity) and its surface are by the increase squared (surface area is a 2D quantity).
So, for the ratio V/SA, if you increase the volume by (approximately) r^3 and its area by (approximately) r^2, this ratio grows at (approximately) a rate proportionate to r.
