## Thomas' Calculus 13th Edition

a=d/$\sqrt 3$ surface area=$2d^{2}$ volume=$d^{3}$/3$\sqrt 3$
Applying Pythagorean theorem, edge diagonal,D= $\sqrt 2$ a. Applying Pythagorean theorem again will yield body diagonal,d=$\sqrt 3$a [$d^{2}$=$a^{2}$+($\sqrt 2a$)^2] Now we have a as function of d. surface area=6$a^{2}$ volume=$a^{3}$ Substitute a in terms of d in the above two equations.