Minimum Surface Area (Without Calculus)?

I suppose the problem is to minimize S, given V = const, right?

This is a very well-known problem and the answer is the cylinder, inscribed in a cube, i.e. h = 2r. Indeed imagine a rectangular cuboid, circumscribed around the cylinder in question - the dimensions of the former are 2r, 2r and h, so that the ratio of the volumes of both solids is 4/π - the same as the ratio of their surface areas.

If the dimensions of a rectangular cuboid are x, y and z, surface area s, and volume v, then according means inequality

s = (2xy + 2yz + 2zx) ≥ 3∛(8x²y²z²) = 6∛(v²)

and v = const would imply s to be minimal when x = y = z, i.e. for cube.

By the way, another similar problem is about a right circular cone instead of cylinder. If its volume is constant, the surface area is minimal if the cone is inscribed in a regular tetrahedron.

V = pi*r^2*h = 32 h = 32/(pi*r^2) A = pi*r^2 + 2*pi*r*h Sub for h: A = pi*r^2 + 2*pi*r*32/pi*r^2 = pi*r^2 + sixty 4/r = pi*r^2 + sixty 4*r^-a million the speed of change of the realm with appreciate to r is dA/dr = 2*pi*r - sixty 4 The max and min is given even as dA/dr = 0 2*pi*r - sixty 4 = 0 r = sixty 4/2*pi = 32/pi Then h = 32/(pi*r^2) = 32/(pi*(32/pi)^2) = pi/32