Cylinders - how many can be formed?

**ReneePatt** · February 8th 2010, 08:45 AM

The question is: How many 1,000 cubic-inch cylindrical tin cans can be formed from a sheet of metal one yard wide and two yards longs?

I have no idea where to start with this one.

I have the surface area of a cylinder formula (2*pi*r*h) and the volume of a cylinder (pi*r^2*h) but can't figure out how that will help me in solving this problem.

**Archie Meade** · February 8th 2010, 09:09 AM

Originally Posted by ReneePatt

The question is: How many 1,000 cubic-inch cylindrical tin cans can be formed from a sheet of metal one yard wide and two yards longs?

I have no idea where to start with this one.

I have the surface area of a cylinder formula (2*pi*r*h) and the volume of a cylinder (pi*r^2*h) but can't figure out how that will help me in solving this problem.

It depends whether you are going to be making the circular top and bottoms from the same piece of metal.

Probably you just need to work out how many curved surfaces for the cylinder you can cut from the strip.

Basically, you must convert yards to inches.

As 1 yard = 3 feet
3 feet = 12 inches,

there are 36 inches in a yard.

The strip is 36 by 72 inches in area.

You must find the surface area of a 1000 cubic inch volume cylinder.

If you allow the height of the cylinders to be 1 yard = 36 inches,
then you can work out the cylinder radius from the volume formula.
Then you can use the surface area formula to discover the curved surface area of the "1000 cubic inch"-volume cylinder.

As you are not given height or radius of the cylinder,
it's convenient to choose the height to equal one side of the strip.

**Soroban** · February 8th 2010, 02:04 PM

Hello, ReneePatt!

This is a messy problem . . .
I assume you are allowed Calculus.

How many 1,000 cubic-inch cylindrical tin cans can be formed
from a sheet of metal one yard wide and two yards long?

We have: . $36 \times 72 \:=\:2592\text{ in}^2$ of sheet metal.

The volume of one cylinder is 1000 in³: . $\pi r^2h \,=\,1000 \quad\Rightarrow\quad h \,=\,\frac{1000}{\pi r^2}$ .[1]

The surface area of a cylinder (including the top and bottom) is:
. . . . $A \:=\:2\pi r^2 + 2\pi rh$ .[2]

Substitute [1] into [2]: . $A \;=\;2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right) \:=\:\frac{2(\pi r^3 + 1000)}{r}$

This is the amount of sheet metal used by one can.

We have 2592 in² of sheet metal availalble.

The number of cans that can be made is: . $N \;=\;\frac{2592}{\frac{2(\pi r^3+1000)}{r}} \;=\;\frac{1296r}{\pi r^3 + 1000}$ .[3]

Differentiate: . $N' \;=\;1296\left[\frac{(\pi r^3+1000)\cdot1 - r(3\pi r^2)}{(\pi r^3+1000)^2}\right] \;=\;1296\left[\frac{1000-2\pi r^3}{(\pi r^3+1000)^2}\right]$

Equate to zero: / $1296\left[\frac{1000-2\pi r^3}{(\pi r^3+1000)^2}\right] \;=\;0 \quad\Rightarrow\quad 1000-2\pi r^3\:=\:0 \quad\Rightarrow\quad r^3 \:=\:\frac{500}{\pi}$

. . Hence: . $r \:=\:\sqrt[3]{\frac{500}{\pi}}$

Substitute into [3]: . $N \;=\;\frac{1296\left(\sqrt[3]{\frac{500}{\pi}}\right)}{\pi\left(\frac{500}{\pi} \right) + 1000} \;=\; \frac{108}{125}\sqrt[3]{\frac{500}{\pi}} \;=\;4.682241246$

Therefore, four cans can be made.

**ReneePatt** · February 9th 2010, 02:28 PM

Wow! That's intense. I would have never figured that out. Thanks so much for your help.

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