OFFSET
1,3
COMMENTS
A square pyramid with a height of h and a base of size s X s has volume V = (1/3)*s^2*h, so a square pyramid of unit volume has s = sqrt(3/h), and the slant height of each of the four triangular faces is t = sqrt(h^2 + (s/2)^2) = sqrt(h^2 + 3/(4*h)), and the total area of the four faces is A = 4*(s*t/2) = sqrt(12*h^3 + 9)/h; this area is minimized at h = (3/2)^(1/3), where it reaches A = 3^(7/6)*2^(1/3).
If the total surface area of all five faces including the square base is to be minimized, then the resulting height is 6^(1/3) (cf. A005486). - Jon E. Schoenfield, Nov 11 2018
LINKS
Eric Weisstein's World of Mathematics, Pyramid.
FORMULA
Equals (3/2)^(1/3) = (1/2)*A010584.
EXAMPLE
1.14471424255333186780804221193967700891590692078793...
MATHEMATICA
RealDigits[Surd[3/2, 3], 10, 120][[1]] (* Amiram Eldar, Jun 21 2023 *)
PROG
(PARI) sqrtn(3/2, 3) \\ Michel Marcus, Oct 23 2018
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Jon E. Schoenfield, Oct 22 2018
STATUS
approved