# A water cup in the shape of a right circular cone is to be constructed!!!A water cup in the shape of a right circular cone is to be constructed by removing the circular sector from a circular...

A water cup in the shape of a right circular cone is to be constructed!!!

A water cup in the shape of a right circular cone is to be constructed by removing the circular sector from a circular sheet of paper of radius a and then joining the two straight edges of the remaining paper. Find the dimensions of the cup with the largest volume that can be constructed.

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Given the dimensions above:

(1)The volume of a cone is given by `V=pi/3r^2h` , where `r` is the radius of the base and `h` is the height of the cone. With this construction, `a` will be the slant height of the cone (The length of a segment drawn from the apex to the base along the side of the cone).

Then `r^2+h^2=a^2 => h=sqrt(a^2-r^2)` .

(2)We can rewrite the formula for the volume as `V=pi/3r^2sqrt(a^2-r^2)` .

(3) To maximize the volume, we find the first derivative of the volume function; setting the derivative equal to zero yields the critical values from which a maximum must come.

(4) Noting that `a` is a constant :

`(dV)/(dr)=(2pi r)/3(a^2-r^2)^(1/2)+(pi r^2)/3(1/2)(a^2-r^2)^(-1/2)(-2r)`

`=(2pi r)/3(a^2-r^2)^(1/2)-(pi r^3)/3(a^2-r^2)^(-1/2)`

`= (a^2-r^2)^(-1/2)[(2pi r)/3(a^2-r^2)-(pi r^3)/3]`

`=((pi r)/3(2a^2-3r^2))/sqrt(a^2-r^2)`

Note that `a != r` as this results in the degenerate cone of height zero.

Setting `(dV)/(dr)=0` we have `(pi r)/3=0` or `2a^2-3r^2=0` .

The first case is degenerate, so `3r^2=2a^2` or `r=(asqrt(6))/3` . Substituting for r we get `h=(asqrt(3))/3` .

**Thus the dimensions, given a circle of radius** `a` are `r=(asqrt(6))/3,h=(asqrt(3))/3`