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## A cylindrical canister in a rectangular box- volume

tagged by: pareekbharat86

This topic has 3 expert replies and 6 member replies
pareekbharat86 Master | Next Rank: 500 Posts
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#### A cylindrical canister in a rectangular box- volume

Wed Nov 13, 2013 8:54 am
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

OA is B

My method:

Volume of cylinder is Pi*r*r*h. So we need to know the combination of 2 of 6,8,&10 which will yield the maximum r*r*h.

The following are the possible options-
6/2 as radius and 8 inches height (3,3,8)- 72
Similarly,
3,3,10- 90
OR 4,4,10- 160
OR 5,5,8- 200

Therefore i feel answer should be C

_________________
Thanks,
Bharat.

hotcool030 Newbie | Next Rank: 10 Posts
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Sun Jul 02, 2017 4:59 am
What about the case of 10 as base and 8 as height
then the volume is 200pi know, in that case radius is 5.

theCodeToGMAT wrote:
Another approach:

Volume = pi * (r)^2 * h
So, lets consider cases
10x8 = (4)^2 * 6 = 96
10x6 = (3)^2 * 8 = 72
8x6 = (3)^2 * 10 = 90
So, 10x8 is the dimension which results in maximum volume

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Sun Jul 02, 2017 9:29 am
Hi hotcool030,

You have to be careful about how you approach the math here. Remember that the base of a cylinder has uniform length and width (re: the diameter of the base), so if the 'height' of the rectangular solid is 8, then the 'base' of the rectangular solid is 10 x 6. This means that the widest cylinder that could fit that space would have a diameter of 6 (NOT 10) and a radius of 3.

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Brent@GMATPrepNow GMAT Instructor
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Wed Nov 13, 2013 9:05 am
pareekbharat86 wrote:
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

OA is B

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 6x8 side, on the 6x10 side, or on the 8x10 side

If you place the base on the 6x8 side, then the cylinder will have height 10, and the maximum radius of the cylinder will be 3 (i.e., diameter of 6).
So, the volume of this cylinder will be (pi)(3Â²)(10), which equals 90(pi)

If you place the base on the 6X10 side, then the cylinder will have height 8, and the maximum radius of the cylinder will be 3 (i.e., diameter of 6).
So, the volume of this cylinder will be (pi)(3Â²)(8), which equals 72(pi)

If you place the base on the 8x10 side, then the cylinder will have height 6, and the maximum radius of the cylinder will be 4 (i.e., diameter of 8).
So, the volume of this cylinder will be (pi)(4Â²)(6), which equals 96(pi)

So, the greatest possible volume is 96(pi) and this occurs when the radius is 4

Cheers,
Brent

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Brent@GMATPrepNow GMAT Instructor
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Wed Nov 13, 2013 9:11 am
pareekbharat86 wrote:
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

OA is B

My method:

Volume of cylinder is Pi*r*r*h. So we need to know the combination of 2 of 6,8,&10 which will yield the maximum r*r*h.

The following are the possible options-
6/2 as radius and 8 inches height (3,3,8)- 72
Similarly,
3,3,10- 90
OR 4,4,10- 160
OR 5,5,8- 200

Therefore i feel answer should be C
To get a volume of 200(pi), you are saying that the height of the cylinder is 8 and the diameter is 10.
This means that the cylinder's circular base is on the side with dimensions 6x10
A circle placed on the side with dimensions 6x10 cannot have a diameter of 10 (it won't fit)

Cheers,
Brent

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Mathsbuddy Master | Next Rank: 500 Posts
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Wed Nov 13, 2013 9:17 am
While your geometrical maths is good, there's a 'trick' to this question - which is hard to spot without a diagram:

A) The biggest circle that will fit on a 10"x6" face has a radius of 3"
B) The biggest circle that will fit on a 6"x8" face has a radius of 3" too
C) The biggest circle that will fit on a 10"x8" face has a radius of 4"

Therefore, the cylinders have volumes of:

A) 3^2 * 8 * pi = 72pi
B) 3^2 * 10 * pi = 90pi
C) 4^2 * 6 * pi = 96pi

Therefore the cylinder with radius 4 inches yields the largest volume!

theCodeToGMAT Legendary Member
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Wed Nov 13, 2013 9:37 am
Another approach:

Volume = pi * (r)^2 * h
So, lets consider cases
10x8 = (4)^2 * 6 = 96
10x6 = (3)^2 * 8 = 72
8x6 = (3)^2 * 10 = 90
So, 10x8 is the dimension which results in maximum volume
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Last edited by theCodeToGMAT on Sun Nov 17, 2013 9:22 am; edited 1 time in total

Mathsbuddy Master | Next Rank: 500 Posts
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Sun Nov 17, 2013 9:10 am
theCodeToGMAT wrote:
Another approach:
True that this gives you the biggest radius, but the question wants the biggest volume.
It is only when each radius squared is multiplied by the corresponding length that we are convinced that this biggest radius option yields the biggest volume.

theCodeToGMAT Legendary Member
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Sun Nov 17, 2013 9:18 am
Yep, actually I drew what you had explained in post above but I mistakenly skipped to include that calculation

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gopalsbhati Newbie | Next Rank: 10 Posts
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Sun Oct 29, 2017 12:32 am
if i try to solve in this fashion-
we have to maximize volume of cylindrical canister subjected to r+h=14 as while scanning the 6,8,10 units this r+h will be maximum

volume V=pi*r*r*h -k(r+h-14)
maximize it and equal to 0
differentiating wrt to r,h and k
eq1 pi*2r*h=k
eq 2 pi*r*r=k
eq 3 r+h=14
solving 1&2 r=2h and from eq 3
h=14/3 and r=28/3
whats wrong in this approach

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