Manhattan Circles Problem Set 2

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by Brent@GMATPrepNow » Mon Nov 26, 2012 7:13 am
theachiever wrote:A rectangular box has the dimensions 12inches X 10inches X 8 inches.What is the largest possible volume of a right cylinder that is placed inside the box?
There are 3 different ways to position the cylinder (with the base on a different side each time).
You can place the base on the 12x10 side, on the 10x8 side, or on the 12x8 side

If you place the base on the 12x10 side, then the cylinder will have height 8, and the maximum radius of the cylinder will be 5 (i.e., diameter of 10).
So, the volume of this cylinder will be (pi)(5^2)(8), which equals 200(pi)

If you place the base on the 10x8 side, then the cylinder will have height 12, and the maximum radius of the cylinder will be 4 (i.e., diameter of 8).
So, the volume of this cylinder will be (pi)(4^2)(12), which equals 192(pi)

If you place the base on the 12x8 side, then the cylinder will have height 10, and the maximum radius of the cylinder will be 4 (i.e., diameter of 8).
So, the volume of this cylinder will be (pi)(4^2)(10), which equals 160(pi)

So, the greatest possible volume is [spoiler]200pi[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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