Question

How do I solve this?

You are given a circular piece of paper with a radius of R inches. Your first task is find the central angle of the sector that should be removed to make a cone with the largest volume when the remaining part of the circle is manipulated to forma a cone. (This was demonstrated in class a few weeks ago.) Your second task is to find the rate of change of the volume of the cone when the radius of the cone is half of R and the circumference is decreasing by 0.3 inches per minute.

Answer 1

Part 1
We are given that R is the radius of the circular piece of paper. We are going to cut a wedge with angle `theta` to form a cone.

Let `r =` the radius of the cone
Let `h =` the height of the cone

The volume of the cone is:

`V = 1/3pir^2h" [1]"`

The Pythagorean theorem gives us the relationship of r and h to R:

`R^2 = r^2+h^2`

Solve for `r^2`:

`r^2 = R^2-h^2" [2]"`

Substitute equation [2] into equation [1]:

`V = 1/3pi(R^2-h^2)h`

Distribute the h:

`V = 1/3pi(R^2h-h^3)`

Compute the derivative with respect to h:

`(dV)/(dh) = 1/3pi(R^2-3h^2)`

To find the maximum, set the first derivative equal to 0 and solve for h:

`0 = 1/3pi(R^2-3h^2)`

`h^2 = 1/3R^2`

`h = sqrt3/3R`

To find the value of `r^2` substitute `h^2 = 1/3R^2` into equation [2]:

`r^2 = R^2-1/3R^2`

`r^2 = 2/3R^2`

To obtain the maximum volume, substitute the values for `r^2` and h into equation [1]:

`V = 1/3pi(2/3R^2)(sqrt3/3R)`

Simplify:

`V_"max" = (2sqrt3)/27piR^3`

We need to find the central angle `theta`

The circumference of the base of the cone is:

`C_"base" = 2pir`

The circumference of the circle is:

`C_"circle" = 2piR`

The arclength of the wedge is:

`S = Rtheta`

We can write:

`S = C_"circle"-C_"base"`

Substituting in the equivalents:

`Rtheta = 2piR-2pir`

We know that `r^2 = 2/3R^2` so we can substitute `r = sqrt(2/3)R`

`Rtheta = 2piR-2pisqrt(2/3)R`

`theta = 2pi((3-sqrt6)/3)`

Part 2
We want to fine the rate of change of the volume, `(dV)/(dt)`

We are given `(dC_"base")/(dt) = -0.3" in/min"`

It follows that `(dr)/(dt) = -0.3/(2pi)" in/min"`

Using the chain rule we see that we need (dV)/(dr) evaluated at `r = R/2`:

`(dV)/(dt)= (dV)/(dr)|_(r=R/2)(dr)/(dt)`

Compute `(dV)/(dr)`

Start with:

`V = 1/3pir^2h" [1]"`

Substitute `h = sqrt(R^2-r^2)`

`V = 1/3pir^2sqrt(R^2-r^2)" [1.1]"`

`(dV)/(dr) = 1/3pi(2rsqrt(R^2-r^2)-(r^3)/sqrt(R^2-r^2))`

Evaluate at `r = R/2`:

`(dV)/(dr)|_(r=R/2) = 1/3pi(Rsqrt(R^2-R^2/4)-(R^3/8)/sqrt(R^2-R^2/4))`

`(dV)/(dt)= 1/3pi(Rsqrt(R^2-R^2/4)-(R^3/8)/sqrt(R^2-R^2/4))(-0.3/(2pi))" in/min"`

I am getting warnings about the length of this answer so I will leave the simplification to you.