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Is there a faster way?

Expert replies
by alex.gellatly » Fri Sep 07, 2012 7:47 pm
The inside dimensions of a rectangular wooden box are 6 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?

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I know how do solve this problem using the "OG way"(testing all the ways) , but it takes way too long. Is there a faster approach that I'm missing? Thanks!
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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Source: — Problem Solving |

by Anurag@Gurome » Fri Sep 07, 2012 8:04 pm
alex.gellatly wrote:The inside dimensions of a rectangular wooden box are 6 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?

3
4
5
6
8

I know how do solve this problem using the "OG way"(testing all the ways) , but it takes way too long. Is there a faster approach that I'm missing? Thanks!
I think you missed one of the dimensions: it should be 6 inches by 8 inches by 10 inches. Assuming this, we can solve it like below.

Volume of cylinder = (pi) * radius² * height of cylinder
If the cylinder is placed on 6 in by 8 in. face, then it's maximum radius is 6/2 = 3 and volume will be (pi) * 3² * 10 = 90(pi)
If the cylinder is placed on 6 in by 10 in. face, then it's maximum radius is 6/2 = 3 and volume will be (pi) * 3² * 8 = 72(pi)
If the cylinder is placed on 8 in by 10 in. face, then it's maximum radius is 8/2 = 4 and volume will be (pi) * 4² * 6 = 96(pi)

We can see that the volume is maximum when radius = 4 in.

The correct answer is B.
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by neelgandham » Fri Sep 07, 2012 8:06 pm
Anil Gandham
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