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# Calculus Again

Wed 17 Mar, 2004 07:28 pm

If a container is to hold 8oz.(236.56cm³), what shaped container needs to be constructed to minimize the amount of material needed for construction...a right circular cylinder or a rectangular prism(square base)?

List all calculations so that i will know how to set up this problem.

Thanks
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Brandon9000

1
Thu 18 Mar, 2004 03:07 pm

Express the surface area (amount of material) of a right circular cylinder in terms of the radius and volume (first find the height in terms of the radius and volume). Assume that the volume is constant and differentiate with respect to r, to find the value, as a function of V, that gives the minimum surface area. You now also know the minimum possible surface area for a given volume. Calculate the minimum area for a volume of 236.56cm³.

Now using the same type of logic, find the minimum surface area for a rectangular prism for that same volume.

Which of these two volumes is less?
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g day

1
Thu 25 Mar, 2004 05:17 am
The right square cylinder would have to hold more water for the same surface area then the most optimal rectulangar prism (which is a cube). Circular shapes or spheres tend to hold more volume per surface area than squares or cubes.

Just set the surface area for the two shapes to be equal and see which has the greater volume.

* * *

The most efficent shape for a rectulangar prism (maximum volume for minimum surface area) is a cube - I'll let you prove this for yourself!

A cube's Surface Area = 6 * s ^ 2, where s = length of a side
A cube's Volume = s ^ 3.

So if s ^ 3 = 236.56, then s = 6.185 cm (approx)
Such a cube's S.A. is approx 229.5 cm^2

* * *

The most efficent shape for a cylinder is one where diameter equals height (proof not provided), so if the clyinder is of radius r it must be of height h = 2r

So the Volume of this shape = 2 pi r ^ 3 and SA = 6 pi r ^ 2

V = (pi * r ^ 2) * h = (pi * r ^2) * 2r = 2 pi r ^ 3
SA = 2 * (pi * r ^ 2) + [ (2 * pi * r) * h] = 2 * pi * r * ( r + h) = 2 * pi * r ( r + 2r) = 6 pi r ^ 2

So for Vol = 236.56 then r = 3.35 (approx) and SA must therefore be 211 cm ^ 2

* * *

To work out the shapes I gave you were the most efficient you need either a strong knowledge of geometry or express the formulea as a function of height, width and depth and maximise V whilst minimising SA and same for the cylinder. I knew the answers so I skipped this part of the proof!
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