MSeaV has hired you to design a new prototype for an environmentally friendly plastic water bottle. The water bottle is to have a volume of 750 cm3. You are to consider a cylindrical model that is topped by a cone. The design would have a snip-off top so no cap would be needed. For ergonomic purposes, the radius should be at least 3 cm and at most 6 cm, and the height of the conical piece should be equivalent to its radius. Determine the dimensions and resulting surface area for a cone topped cylindrical plastic water bottle that minimizes the amount of plastic used. Provide a solution to this problem that includes the following: evidence of a plan to solve an optimization problem that includes an algebraic solution using calculus techniques (differentiation, solving for extreme values, testing interval endpoints, etc.); at least one other method to verify your result (graphical or numeric); supporting sketches; and a reflection on the reasonableness of your answer with justification.

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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...

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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.

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