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Solid Geometry - Cuboids and Cylinders

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by nikhil.jejurikar » Fri May 03, 2013 9:31 pm
Whats the best way to solve this?

A crate measures 4 feet by 8 feet by 12 feet on the inside. A stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides. What is the radius, in feet, of the pillar with the largest volume that could still fit in the crate?

A. 2
B. 4
C. 6
D. 8
E. 12
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by Atekihcan » Fri May 03, 2013 10:10 pm
Let us assume the radius of the circular base and height of the cylinder are r and h respectively, the volume of the cylinder is πr²h

The rectangular box has 3 faces : 4x8, 4x12, and 8x12
The circular base of the cylinder must lie on one of these faces.
In each case, the diameter of the circular base cannot be greater than the smaller side of the face.
So, possible value of r are : 4/2, 4/2, and 8/2 ---> 2, 2, and 4

Now, the problem can be solved in two ways.
Algebra:
Maximizing the volume means maximizing r²h.
Now, as r is squared and all possible values of r are greater than 1, maximizing r will have a greater effect in maximizing the volume than maximizing h.
Maximum possible value of r is 4.

Answer : B


Plug the options: Clearly only possible options are A and B.
A. If r = 2, we need to maximize h to maximize the volume. h can be maximum, i.e. 12 if the circular face lies on 4x8 face. Hence, maximum volume = πr²h = π*(2²)*12 = 48π
B. If r = 4, the circular face lies on 8x12 face. So, height of the cylinder = 4. Hence, maximum volume = πr²h = π*(4²)*4 = 64π

As option B gives a larger volume, B is the correct answer.
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by Brent@GMATPrepNow » Sat May 04, 2013 5:41 am
nikhil.jejurikar wrote:Whats the best way to solve this?

A crate measures 4 feet by 8 feet by 12 feet on the inside. A stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides. What is the radius, in feet, of the pillar with the largest volume that could still fit in the crate?

A. 2
B. 4
C. 6
D. 8
E. 12
Volume of cylinder = pi(radius^2)(height)

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 4x8 side, on the 4x12 side, or on the 8x12 side

If you place the base on the 4x8 side, then the cylinder will have height 12, and the maximum radius of the cylinder will be 2 (i.e., diameter of 4).
So, the volume of this cylinder will be (pi)(2^2)(12), which equals 48(pi)

If you place the base on the 4x12 side, then the cylinder will have height 8, and the maximum radius of the cylinder will be 2 (i.e., diameter of 4).
So, the volume of this cylinder will be (pi)(2^2)(8), which equals 32(pi)

If you place the base on the 8x12 side, then the cylinder will have height 4, 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)(4), which equals 64(pi)

So, the greatest possible volume is 64(pi) and this occurs when the radius is 4

Answer = B

If anyone would like additional practice, here are 2 similar questions:
- https://www.beatthegmat.com/og-12-189-pr ... 47627.html
- https://www.beatthegmat.com/manhattan-ci ... 46689.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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