We have a cylinder with a cone on top. The height of the cone is the same as the radius of the cylinder. The radius is restricted from `3<=r<=6.` The volume of this container is 750 cc.

We want to minimize the amount of plastic required to make this container—in other words, we need to minimize the surface area of the container.

If we let r be the radius of the cylinder (and by extension the height of the cone) and let h be the height of the cylinder we have:

`V_(" total" )=V_( "cylinder" )+V_( "cone" )=pi r^2 h + 1/3 pi r^3`

The total surface area is:

`SA_"total"=A_"bottom"+A_"side"+LA_"cone"`

`SA=pi r^2 + 2 pi r h + pi r sqrt(r^2+r^2)` (since height of cone =r)

Now we have a function in two variables h and r. We would like to write this as a function of one variable, so we solve the volume equation for h:

`750=pi r^2 h + pi/3r^3`

`pi r^2 h = 750 - pi/3 r^3`

`h=(750-pi/3 r^3)/(pi r^2)`

Substituting this expression for h into the equation for the total surface area we get:

`SA=pi r^2+2 pi r ((750-pi/3 r^3)/(pi r^2))+pi r sqrt(2r^2)`

Simplifying we get:

`SA=pi r^2+1500/r-pi/3 r^2+sqrt(2) pi r^2`

`=pi r^2(sqrt(2)+2/3)+1500/r`

Now to minimize we can take the derivative with respect to r and find the critical values.

`(d(SA))/(dr)=-1500/r^2+2pi(sqrt(2)+2/3)r`

Setting this equal to zero to find the critical points we get:

`1500=2pi(sqrt(2)+2/3)r^3 ==> r^3=1500/(2pi(sqrt(2)+2/3))`

So `r~~4.859, h~~8.491, SA~~463.05`

See the attachment for the relevant part of the graph of the surface area function.

Note that r is in the bounds required.

**Further Reading**