# Graphical relationship of Surface Area against Volume with Platonic solids

What would be the best way to find the function of the surface area based on the volume for platonic solids, so that the Surface Area to Volume ratio would be comparable as shown in the graph below.

I found this image in Wikipedia but wasn't able to find the mathematics behind in, I'm assuming it's correct.

• for the cube, surface area $A$ is 6(side)$^2$ and volume $V$ is (side)$^3$, so $A=6V^{2/3}$; your picture shows a sphere besides Platonic solids Commented Aug 29, 2019 at 21:27
• The formulas for surfaces and areas of the solids are in the same wikipedia' article, paragraph Mathematical examples Commented Aug 29, 2019 at 21:54

The formula for area will be some constant times the square of the length measurement, so if we use $$L$$ for the length measurement the formula will be like $$A = A_1 L^2.\tag1$$ The formula for volume, on the other hand, is a constant times the cube of the length measurement, like $$V = V_1 L^3.\tag2$$
We can solve for $$L$$ in Equation $$(2)$$ to find that $$L = \sqrt[3]{\frac V{V_1}},$$ and plugging this value of $$L$$ into Equation $$(1)$$ we have $$A = A_1 \left(\sqrt[3]{\frac V{V_1}}\right)^2 = \frac{A_1}{V_1^{2/3}} V^{2/3} .$$
(The notation $$x^{2/3}$$ gives the same result as $$\left(\sqrt[3]{x} \right)^2.$$)
The constant factor $$\frac{A_1}{V_1^{2/3}}$$ is different for each type of polyhedron, but you only need to work it out once for each kind. The measurement $$L$$ does not need to be an edge length; it could be the radius of the inscribed sphere, the radius of the circumscribed sphere, or even the edge of a cube that the polyhedron has been cleverly fitted inside. It just needs to measure the size of the polyhedron somehow, and it needs to be the same measurement in both the area formula and the volume formula.
Once you have worked out the factor, you will have formula for $$A$$ as a function of $$V$$ which you can then use to plot the curves you saw.