What dimensions of a cone minimize surface area when the volume of said cone is 8 cubic inches?

Question

Volume of cone=8in^3
SA of a cone is pi*radius*sqr(radius^2+height^2)
Volume of a cone= 1/3(pi)(radius^2)(height)

Answer 1

Let r denote radius, h denote height

We're told that (1/3)πr²h = 8. Solve this for h: h = 24/(πr²)

Substitute this for h in the expression for surface area. This gives you SA as a function of r. Minimize as usual.

I get r ≈ 1.755, h ≈ 2.481 (assuming that "surface area" means just the lateral area, and not including the base)

Answer 2

Sa=2πr*slant heighrt+πr^2=(2sqrt(r^2+h^2) +r)πr
H interms of slant height =
h=Sqrt(r^2+h^2)*sinθ
plug into volume formula
1/3πr^2sqrt(r^2+h^2)sinθ=8. Divide by r^3
1/3πsqrt(r^2+h^2)/rsinθ=8/r^3

1/3πtanθ. Take r as a constant and differentiate wrt θ
=1/3πsec^2θ=0=π/3(r^2+h^2); r=h minimizes SA
1/3πr^3=8, r=h=(24/π)^.333333=1.97in