Please reference the illustration at screenshots<dot>firefox<dot>com/Cl7zsk8ZAmzCoMDM/brilliant<dot>org to answer the following question:

Question

The Pharaoh would like to cover the lateral sides of his square pyramid with the finest marble. In order to save the cost of construction, the builder tries to reduce the exposed surface area of the pyramid as much as possible.

What is the angle of inclination Θ(in degrees) that minimizes the exposed surface area given a constant volume of the pyramid?

Answer 1

Ok, I saw you diagram, a pyramid with a square base of sides a, and height h, with h1 as the height of each of the 4 triangles.

given :
(1/3) a^2 h = V, where V is a constant
h = 3V/a^2

SA = total surface area covered by marble
= 4(1/2) a h1 = 2(a)(h1)

I can also see that : h1^2 = a^2 /4 + h^2
h1^2 = a^2 /4 + 9V^2/a^4
= (a^6 + 36V^2)/(4a^4)
h1 = (a^6 + 36V^2)^(1/2) / (2a^2)

SA = 2a (a^6 + 36V^2)^(1/2) /(2a^2)
= (a^6 + 36V^2)^(1/2) /a

getting there ...... take d(SA)/da by quotient rule:
SA' = (a (1/2)(a^6 + 36v^2)^(-1/2)(6a^5) - (a^6 + 36V^2)(1) )/a^2
= 0
3a^6/√(a^6 + 36V^2) = √(a^6 + 36V^2)
3a^6 = a^6 + 36V^2
2a^6 = 36V^2
a^6 = 18V^2
a^3 = 3√2 V
V = a^3/ 3√2 BUT from above V = a^2 h/3

a^3/3√2 = a^2 h/3

a/√2 = h

and now for the finale .... drumroll -------

tanØ = h/(a/2) = 2h/a
= 2(a/√2) / a
= 2/√2 or √2

Ø = appr 54.7°

I sure hope I didn't make some silly arithmetic error