Conical paper cups are usually made so that depth is sqrt(2) times radius of rim. Show that the design uses least amount of paper per unit volume.
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.