Save the Snow (Snowy Shenanigans I)

Let $V$ be the amount of snow each child starts with, and $S$ be the surface area in contact with the air. Every child starts with the same amount of snow, and the snow sculpture with the least amount of surface area in contact with the air will have the most snow the next day, so we want to find the minimum $S$ for a constant $V$ .

Fraser leaves the sculpture as a snowman. Let the snowman be constructed by $n$ number of spheres. The volume of one sphere is $V = \frac{4}{3}\pi r^3$ , so the volume distributed over $n$ spheres would be $\frac{V}{n} = \frac{4}{3}\pi r^3$ , which means each sphere would have a radius of $r = \sqrt[3]{\frac{3V}{4n\pi}}$ . The surface area of one sphere in contact with the air is $S = 4 \pi r^2$ , so $n$ spheres with the above $r$ would have a surface area of $S = 4n\pi\sqrt[3]{\frac{3}{4n\pi}}^2\sqrt[3]{V}^2$ . As $n$ increases, so does $S$ , so the minimum $S$ would be when $n$ is minimum, which would be $n = 2$ (to still be a recognizable snowman). When $n = 2$ , $S \approx 6.093\sqrt[3]{V}^2$ .

Owen makes the sculpture into a sphere. We can use the same formula for $S$ as Fraser (see above), but with $n = 1$ . Then Owen's sculpture's surface area in contact with the air is $S \approx 4.836\sqrt[3]{V}^2$ .

Georgina makes the sculpture into a hemisphere. The volume of a hemisphere is $V = \frac{2}{3}\pi r^3$ , which means $r = \sqrt[3]{\frac{3V}{2\pi}}$ . To minimize the surface area in contact with the air, the sculpture can be placed with the flat part against the ground, leaving half of the surface area of a sphere, which is $S = 2 \pi r^2$ . Substituting $r$ we have $S = 2\pi\sqrt[3]{\frac{3}{2\pi}}^2\sqrt[3]{V}^2 \approx 3.838\sqrt[3]{V}^2$ .

Paddy makes the sculpture into a cylinder. The volume of a cylinder is $V = \pi r^2h$ , which means $h = \frac{V}{\pi r^2}$ . To minimize the surface area in contact with the air, the sculpture can be placed with the flat part against the ground, leaving part of the surface area of a cylinder without the bottom circle, which is $S = \pi r^2 + 2\pi rh$ . Substituting h, we have $S = \pi r^2 + \frac{2V}{r}$ . The minimum surface area in contact with the air would be when $S' = 0$ , and since $S' = 2\pi r - \frac{2V}{r^2}$ , this occurs when $r = \sqrt[3]{\frac{V}{\pi}}$ , or when $S = 3\pi\sqrt[3]{\frac{1}{\pi}}^2\sqrt[3]{V}^2 \approx 4.394\sqrt[3]{V}^2$ .

Out of the four children, Georgina 's sculpture has the minimum surface area in contact with the air, and therefore the most snow left for the next day.

The pile of snow will lose heat through its surface. So you would want to minimize the surface to volume ratio of the snow. Since the grass doesn't melt the snow, the shape that minimizes this ratio is the hemisphere, much like a bubble in a bathtub.