# 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.

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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

Also observe:

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

Check!

Note that K doesn't katter in this calculation. We could have set it to 1.