Geometry - Ch.3 (Circles & Cylinders) - In Action #10

[slsu]

Hello,

I had a question regarding the answer to the question below:

10. A rectangular box has the dimensions 12 inches x 19 inches x 8 inches. What is the largest possible volme of a right cylinder that is placed inside the box?[/b][/b]

Answer:

200(3.14) - The radius of the cylinder must be equal to half of the smaller of the 2 dimensions that form the box's bottom. The height, then, can be equal to the remaining dimension of the box. Since the radius is squared in the formula, it is essentially counted twice, while the height is only counted once. Thus, the largest possible radius will result in the cylinder with the largest volume. Therefore, a radius of 5 (half of 10, the smaller of the box's 10 x 12 bottom dimensions) and a height of 8 will result in the largest possible volume:

V = (3.14)(r)(r) x (h) = 25(3.14( x 8 = 200(3.14)

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My questions:
(1) Why do you have to work with the smaller 2 dimensions in order to find the radius?
(2) How does the smaller radius, result in the largest volume? Doesn't the larger radius (as cited in the answer) lead to largest volume?

[StaceyKoprince]

Is there a typo in your problem? You list the dimensions as 12 x 19 x 8, but the explanation seems to indicate the dimensions are 12 x 10 x 8. I'm going to assume that the latter represents the correct dimensions.

Go get a box before you read the below. Seriously. :)

First you ask "Why do you have to work with the smaller 2 dimensions in order to find the radius? "

Technically, you do not work with the smaller 2 dimensions (that would mean using 10 and 8, given our current dimensions). You work with the smaller OF the 2 dimensions you're using. Because we want to maximize volume, we actually want to use the two larger dimensions (12 and 10) to set the radius, because the radius gets squared in the volume formula.

A cylinder has a constant radius; look in your box and imagine sticking a cylinder inside. Whatever you choose as the bottom of your box, you have to map the cylinder to the smaller of the two dimensions, because if you map it to the larger of the two, your cylinder will end up going outside of the box. For instance, if the bottom is 6 x 20 and you tried to make the radius half of 20 (or 10) - you can see that this wouldn't work for the side that is only 6. The cylinder wouldn't fit inside.

So, you have to base the radius on the smaller of whatever two dimensions you have.

Second question: How does the smaller radius, result in the largest volume? Doesn't the larger radius (as cited in the answer) lead to largest volume?

As noted above, the cylinder is supposed to be INSIDE the box. Yes, a larger radius would mean a larger cylinder, but we can't choose a radius so large that the cylinder won't fit in the box - we have to meet that constraint.

We maximize the radius by, first, choosing the two larger dimensions to be the bottom of the box and, second, setting the radius to half of the smaller of those two dimensions.

[slsu]

Thank you Stacey!

You were correct - the 19 was a typo. Should have been 10.

And yes, I have a box sitting right in front of me. It was much easier to visually see how you derived the answer!

[gmat703]

I have a question,

I understand why the radius cannot be the larger dimension.
But what is wrong in selecting the height as the larger dimension.

Result: r=5, h=12
Volume(cylinder): 5*5*12*Ï€=300Ï€.

[messi10]

I have a question,

I understand why the radius cannot be the larger dimension.
But what is wrong in selecting the height as the larger dimension.

Result: r=5, h=12
Volume(cylinder): 5*5*12*Ï€=300Ï€.

That cylinder wont fit in the box. If the radius is 5 then the diameter is 10. You are trying to place this on the side with dimensions 10 x 8. This is not possible.

[RonPurewal]

I have a question,

I understand why the radius cannot be the larger dimension.
But what is wrong in selecting the height as the larger dimension.

Result: r=5, h=12
Volume(cylinder): 5*5*12*Ï€=300Ï€.

That cylinder wont fit in the box. If the radius is 5 then the diameter is 10. You are trying to place this on the side with dimensions 10 x 8. This is not possible.

yep. in that case you'd have r = 4.