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
8/2 as radius 4,4,6- 96
OR 4,4,10- 160
10/2 as radius 5,5,6- 150
OR 5,5,8- 200
Therefore i feel answer should be C
A cylindrical canister in a rectangular box- volume
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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
Volume of cylinder = pi(radius²)(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 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)(3²)(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)(3²)(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)(4²)(6), which equals 96(pi)
So, the greatest possible volume is 96(pi) and this occurs when the radius is 4
Answer: B
Cheers,
Brent
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To get a volume of 200(pi), you are saying that the height of the cylinder is 8 and the diameter is 10.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
8/2 as radius 4,4,6- 96
OR 4,4,10- 160
10/2 as radius 5,5,6- 150
OR 5,5,8- 200
Therefore i feel answer should be C
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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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!
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!
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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
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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True that this gives you the biggest radius, but the question wants the biggest volume.theCodeToGMAT wrote:Another approach:
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.
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Yep, actually I drew what you had explained in post above but I mistakenly skipped to include that calculation
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What about the case of 10 as base and 8 as height
then the volume is 200pi know, in that case radius is 5.
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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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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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.
GMAT assassins aren't born, they're made,
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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
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