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31 Oct 2010, 08:25
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:18) correct
42%
(02:04)
wrong
based on 116
sessions
History
Date
Time
Result
Not Attempted Yet
A product designer is trying to design the largest container possible hat will fit into a 6x12x14 box. Because of constraints in the manufacturing process, he must make the container in the shape of a cylinder, which will then be placed base down inside the box. What is the volume of the largest cylinder that can fit in the box?
A) 108π
B) 126π
C) 216π
D) 504π
E) 864π
A) 108π
B) 126π
C) 216π
D) 504π
E) 864π
31 Oct 2010, 11:21
Kudos
Bookmarks
Jumphi97
Volume of a cylinder is \(\pi r^2 h\)
There is 3 configurations possible :
Height along side 6 --> Base on area of 12x14 --> Max radius is 6 : Vol = \(216 \pi\)
Hieght along side 12 --> Base on area of 6x14 --> Max radius is 3 : Vol = \(108 \pi\)
Height along side 14 --> Base on area of 6x12 --> Max radius is 3 : Vol = \(126 \pi\)
So max volume possible is (C)
31 Oct 2010, 11:34
Kudos
Bookmarks
Volume of a cylinder = \(\pi r^2 h\)
Since the cylinder is placed into the box base down, we have two options for the base of the cylinder: 6 or 12. For whichever two numbers we choose as the base of the box, the diameter of the cylinder has to be equal to the lesser of the two dimensions in order to fit.
Our goal is to maximize the volume of the cylinder, which means we want the largest possible base. This is because \(r^2\) will create a larger number than \(h\).
The base of the box would therefore be 12x14 with a height of 6. This allows for the following dimensions of the cylinder:
\(\pi 12^2 * 6\) = \(216\pi\)
Since the cylinder is placed into the box base down, we have two options for the base of the cylinder: 6 or 12. For whichever two numbers we choose as the base of the box, the diameter of the cylinder has to be equal to the lesser of the two dimensions in order to fit.
Our goal is to maximize the volume of the cylinder, which means we want the largest possible base. This is because \(r^2\) will create a larger number than \(h\).
The base of the box would therefore be 12x14 with a height of 6. This allows for the following dimensions of the cylinder:
\(\pi 12^2 * 6\) = \(216\pi\)
