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Please help with PS question

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by madsport » Sun Sep 07, 2008 6:23 pm
The inside dimensions of a rectangular wooden box are 6 inches by 8 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?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

The answer is B
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by rabab » Sun Sep 07, 2008 7:58 pm
I know!! I am just as stuck as you're with the same question!
Can anyone help pls?
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by ritula » Sun Sep 07, 2008 9:23 pm
Hmmmm. well let me give it a try. But I am not sure if I am doing it the right way.
Lets say the volume is same and height of canister be the same as rectangular box. Then
6*8*10=3.14 *(r^2)*10
Lets approximate 3.14 as 3 to simplify hethings
48/3=r^2
r=4
Hence B
Philosophers have interpreted world in various ways, the point is to change it!
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by mankuthimma » Mon Sep 08, 2008 12:16 am
Here is how I solved the problem.

We have to find the combination of radius and height that results in maximum volume. The base of the rectangular box can be 6x8, 6x10 and 8x10.
If the base is 6x8, then the circular base of a cylinder placed inside the rectangular box can have the diamter 6. That means, radius is 3 and the volume 90(pi).
Similarly, if the base is 6x10, the radius is again 3 and the volume is 72(pi).
If the base is 8x10, then the radius is 4 and the volume is 96(pi). So, the maximum volume is 96(pi) and the maximum radius is 4.

Answer is B.
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by rabab » Mon Sep 08, 2008 12:38 am
mankuthimma wrote:. The base of the rectangular box can be 6x8, 6x10 and 8x10.
Pardon me if I sound stupid, but why are we considering the base to be 6x8, 6x10 and 8x10 only? Why not 10X6, 10X8 and so on?

Thanks for your reply anyways.
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