To answer this question, you have to optimize the radius subject to some constraint. In this case, you are optimizing the surface area of the cone with a given volume. So, we'll assume that the Volume is fixed and non-zero, of course.
The surface area of a cone (except for its base) is:
S = pi*r*s, where r is the radius and s is the length of a side.
So we assume that the volume is fixed at V_k.
For a cone, we have a volume formula: V = (1/3) pi r^2 h, where h is the height of the cone.
V_k = (1/3) pi r^2 h --> r^2 h = 3V_k/pi = K
h = 3V_k/pi/r^2 --> r^2 = K/h
h^2 + r^2 = s^2
Now we are about ready to minimize our area. Note that minimizing B = S^2 will be easier, so we'll do that.
S = pi*rs
S^2 = B = (pi*r)^2*s^2 = (pi*r)^2*(h^2 + r^2)
B = pi^2*K/h*(h^2 + K/h) = pi^2( K^2/h^2 + K*h )
Differentiate B with respect to h, and set equal to zero to minimize:
D(B) = pi^2( -2K^2/h^3 + K ) = 0
1 - 2K/h^3 = 0
h^3 = 2K
h = (2K)^1/3
Remember r^2 = K/h:
r^2 = K(2K)^-1/3
r = K^(1/2) / (2K)^(1/6)
Now to answer your question:
"the depth is square root of two times radius of rim"
h =? (2)^1/2 r
(2K)^1/3 =? 2^1/2 * K^(1/2) / (2K)^(1/6)
(2K)^1/3 =? 1/(K)^(1/2) * (2K)^(1/2) * K^(1/2) / (2K)^(1/6)
(2K)^1/3 =? (2K)^(1/2) * K^(1/2) / (2K)^(1/6)
(2K)^1/3 == (2K)^1/3
Note that K doesn't katter in this calculation. We could have set it to 1.