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Anonymous Poster

Hot Water Cylinder Differentiation

05/28/2010 5:59 PM

Hi, I need help with a calculus qestion

To reduce heat loss, the surface area of a hot-water tank must be kept to a minimum. If such a tank is 125 litters in capacity, and can be approximated by a cylinder in shape with a hemispherical end cap; calculate the radius and overall height for minimum heat loss.

This is how far I've gotten:

The volume is the sum of the volume of a hemisphere and a cylinder.

The surface area is
=
Isolate h in the Volume equation.


Substitute for h into the Surface area equation.

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#1

Re: hot water cylinder differentiation

05/28/2010 7:52 PM

Don't have much, do ya?

Why not delete the, "cylinder in shape" and use two hemispheres joined at the middle?

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#4
In reply to #1

Re: hot water cylinder differentiation

05/29/2010 5:17 AM

Ya think he might "get" that you're trying to steer him in the right direction, if that's "all he's got?"

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#2

Re: Hot Water Cylinder Differentiation

05/28/2010 8:28 PM

Two geometrical shapes must be taken into consideration. and if there are two end caps it is that of a circle and a cylinder. solve for h in terms of r in the volume equation, and use substitution in the second equation. then solve for r in the surface area equation. find when the first derivative of that equation is equal to 0. then you should find your answer.

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#3

Re: Hot Water Cylinder Differentiation

05/29/2010 12:25 AM

If h is the height of the cylindrical part, then h = 0; if h is the overall height, then h = 2r.

It depends on sorting out that ambiguity.

(This is just another way of saying lynlynch's GA.)

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#5

Re: Hot Water Cylinder Differentiation

07/10/2024 6:49 AM

A sphere about twice the diameter of a soccer football fits the minimum surface-to-volume criterion; no calculus needed.

No mention of insulation or windspeed...

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