# How do you solve the optimization problem(calculus)?

cone paper cups usually made so that the depth is square of two times radius of rim. Show that this design requires the least amount of paper per unit.

tip: solve for h in the volume of a cone, then put that where h is for the Area of the cone.

First you need relations for the volume and area of a cone. The volume is well known:

v = π r^2 h/3

To get the area, consider the pattern when a cone is unwrapped. It is a segment of a circle of radius b. Where:

b = √(h^2 + r^2)

The circumference of the pattern is the cicumference of the assembled cone (2 π r). The circumference of a complete circle is 2 π b so the fraction of the circle is the circumference ratio: 2π r / (2π b) = r/b so the area of the cone is:

A = π r b = π r √(h^2 + r^2)

To find the maximum volume for a given area, substitute the area realtion into the volume relation. To do this, invert the area relation:

A = π r √(h^2 + r^2)

h = √((A/(π r))^2 - r^2)

Substituting into the volume:

v = π r^2 h/3 = π r^2 √((A/(π r))^2 - r^2)/3

Simplify a little:

v = π r^2 √((A/(π r))^2 - r^2)/3 = √(A^2 r^2 - π^2 r^6 )/3

Determine dv/dr and set it to zero to find the r that gives maximum volume:

dv/dr = (1/2)((A^2 r^2 - π^2 r^6 )^(-1/2) * (2rA^2 - 6π^2 r^5)/3

To equal zero, only the variable term in the numberator need be zero:

2rA^2 - 6π^2 r^5 = 0

A^2 = (6π^2 r^5)/(2r) = 3 π^2 r^4

The area formula from above: A = π r √(h^2 + r^2) so:

A^2 = π^2 r^2 (h^2 + r^2)

Equating the two:

3 π^2 r^4 = π^2 r^2 (h^2 + r^2)

3 r^4 = h^2 r^2 + r^4

h = √(2) r

So the height is the square root of 2 times the radius, not the square of 2 times the radius as stated in the question.

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