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wooden box and a cylindrical canister inside with max radius [#permalink]
 17 Mar 2010, 18:30
Question Stats:
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33% (00:58) wrong based on 0 sessions
Can someone show me how to solve the following question?
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
Correct Ans is B.
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
20 Aug 2010, 07:22
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kp1811 wrote: hb05sv wrote: Can someone show me how to solve the following question?
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
Correct Ans is B. let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches) If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches) If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches) so for r = 4inches the cylinder will have maximum area. so B - 4 Sorry guyz but dont understand your ways My approach is; pi approx. = 3,14 The box volume is 480 so; Cylinder max. volume should be pi*5^2*8 = 200pi means more then 600 so cant be, If the radius cant be 5 so it should be 4; pi*4^2*10= slightly more then 480 so CANT BE then the answer is; pi*4^2*6 < 480 it means The radius should be "4". B
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
21 Aug 2010, 16:17
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Since we want Max Volume for cylinder hence Max Vol can only max if the Radius is maximum We can always have Cylinder with radius 3 in this Box but we need max radius if we take (6 and 8) or (6 and 10) or (10 and 8) So let us pick 5 as radius so the diameter will be 10 but other two sides are 8 and 6 which cause a cylinder with 10 radius out of the box. Hence 5 is not the answer Anything above 5 i.e 6 and 8 are gone Now we come to our last option 4 if we take sides 10 and 8 as the base we can surely incorporate cylinder inside the box hence our answer is B i.e. 4. Try to give it a thought because I have not used any calculation to solve this question. And don't forget in Gmat Exam we need to conserve all our energies because after Quants exam the beast awaits......VERBAL!!!!!!
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
17 Mar 2010, 20:02
hb05sv wrote: Can someone show me how to solve the following question?
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
Correct Ans is B. let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches) If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches) If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches) so for r = 4inches the cylinder will have maximum area. so B - 4
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
17 Mar 2010, 20:26
kp1811 wrote: let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches)
If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches)
If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches)
so for r = 4inches the cylinder will have maximum area.
so B - 4 Nice explanation - To be more specific, logically, for any face down of the cube, the smaller length only can be the diameter of the cylindrical canister. So either ways, it is 6 or 8. The height will be either the shortest or the longest dimension accordingly. Volume of cylinder is Pi * r squared * h. The greater the value of r^2*h, the greater the volume.
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
18 Mar 2010, 15:23
BarneyStinson, Can you explain why logically the diameter of the cylinder will be one of the shortest?
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
18 Mar 2010, 17:04
hb05sv wrote: BarneyStinson, Can you explain why logically the diameter of the cylinder will be one of the shortest? Because if the diameter were wider say 8", it won't fit into a box that has one side 6", when the face down is 8" X 6". You should definitely read my blog!!!
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
31 Aug 2010, 11:15
amneetpadda wrote: And don't forget in Gmat Exam we need to conserve all our energies because after Quants exam the beast awaits......VERBAL!!!!!! And remember to reserve some extra energy for AWA:)
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
01 Sep 2010, 00:34
Calculate r^2 * h for all cases you gat the answer
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
01 Jun 2011, 05:08
6*8*10-hence:
side 6*8 r=3 h=10
side 6*10 r=3 h=8
side 8*10 r=4 h=6
pir^2 will be bigger at side 8*10
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
13 Jun 2011, 01:15
l = 10,w = 8 and h = 6 gives max volume. r = 4.
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Re: wooden box and a cylindrical canister inside with max radius [#permalink]
23 Nov 2011, 09:40
The above mentioned solutions are correct, but from my perspective the question could be misleading (at least for non-natives).
The question states "a cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on (ANY) one of its six faces".
Since the canister has a fixed volume and we cannot be sure on which face the wooden box will stand, we would have to assume that it could also stay on the smallest possible area, namely 6 by 8.
If so, the correct solution, 4, would be wrong, since the diameter would exceed the side length 6.
Sorry if I confused you guys, but I personally dislike such questions, since they inhabit the potential for incorrect choices only due to the unclear phrasing of the question.
My 2 cents.
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Re: wooden box and a cylindrical canister inside with max radius
[#permalink]
23 Nov 2011, 09:40
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