what is the max volume of a right circular cone with slant height s?

Answer 1

Normal Cone volume = (1/3)*(PI)*(R)*(R)*H where R=radius of cone and H = hight of cone. If I understand the "right circular cone" correct, the angle made by the tip of the cone is a right angle. This will mean that R=H.

This will give Right Angle Cone Volume = one third x Pi x Radius to the power 3
= (1/3)*(PI)*(R)*(R)*(R) = (1/3)*(PI)*(R)^3 ...........formula A

The relation between s and R can be determined by using
Sin 45 deg = R/s this give R=0,7071s
Substitute in formula A

=(1/3)*(PI)*(0,7071s)^3
=0,3702*(s)^3

Answer 2

There's only one volume for a right circular cone with slant height s, and that's

V = 1/3 pi 2^(-3/2) s^3

Answer 3

V = (1/3)πr^2h
r = ssinθ
h = scosθ
V = (1/3)s^3sin^2θcosθ
dV/dθ = (1/3)s^3 [-sin^3θ + 2sinθcos^2θ] = 0 for max V
sin^3θ = 2sinθcos^2θ
sin^2θ = 2cos^2θ
tan^2θ = 2
tanθ = √2
θ = 54.736°
V = (1/3)s^3(2/3)(1/√3)
V = ((2√3)/27)s^3
V = (2√3)(s/3)^3
or
V = 0.12830006s^3

verifying numerically,
54.70000000 0.128299911
54.73561032 0.12830006
54.80000000 0.128299574