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 containerâ€”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.

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