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The inside dimensions of a retangular wooden box are 6 [#permalink]
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22 Feb 2008, 04:07
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191. The inside dimensions of a retangular 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 maximum volume? A. 3 B.4 C. 5 D. 6. E. 8 Hi friends!, please can help me a shortcut for this? many thanks
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Re: retangular box [#permalink]
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22 Feb 2008, 04:37
I will go for C. for the canister to have max.volume, the max. could be the volume of the box thus volume of the cylinder =volume of the box Pi r^2 h = l * b*h now the height of the canister cannot be more than the height of the box (to fit in) thus Pi r^2 *6= 6*8*10 3.14 r^2 =80 => r^2= 25.47=>r=5.04 or 5 I drew a diagram for visualizing. regards, Pras
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Re: retangular box [#permalink]
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22 Feb 2008, 07:10
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sondenso wrote: 191. The inside dimensions of a retangular 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 maximum volume?
A. 3 B.4 C. 5 D. 6. E. 8
Hi friends!, please can help me a shortcut for this? many thanks B is the answer. Here is an easy way to solve it. there are 3 options: the high of the canister is equal to 10,8 or 6. if it's equal to 10 > the radius is min(8,6)/2 = 3 > volume is 10*pi*3^2 = 90*pi if it's 8 > radius is min(10,6)/2 = 3 > volume obviously below the first one if it's 6 > radius is min(10,8)/2 = 4 > volume is 6*pi*4^2 = 96*pi > radius is 4 > B



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Re: retangular box [#permalink]
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22 Feb 2008, 08:13
My logic 
Diameter will never be greater than 8 because it should be equal to the "shorter" side, so only A or B can be possible answers. Since volume is proportional to d^2*h, just do the math for these options.
B is the answer



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Re: retangular box [#permalink]
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22 Feb 2008, 14:48
Can someone elaborate on this explanation...I dont see how the cyclinder could have a radius greater than 3 if it is to fit into the box
the narrowest point of the box is 6; how could something with a radius of 4 (diameter of 8) fit in there



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Re: retangular box [#permalink]
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22 Feb 2008, 18:27
B it is. Whts the Oa?
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Re: retangular box [#permalink]
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22 Feb 2008, 19:05
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sondenso wrote: 191. The inside dimensions of a retangular 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 maximum volume?
A. 3 B.4 C. 5 D. 6. E. 8
Hi friends!, please can help me a shortcut for this? many thanks Alright, here is my simple and effective way to approach this problem. A cylinder has a base and height. The base is obviously going to rest on one of the many faces of the rectangular box. Lets say it rests on the face with dimensions (6,8) in which case the height of the cylinder will be 10. Remember the face on which the cylinder rests, the radius of the cylinder will be half of the length of the smaller edge of the face. So if the cylinder rests on face (6,8), the radius is going to be 3 inches. Note it cannot be 4 inches because if it is then it will go beyond the boundaries of the (6,8) plane. If you don't agree with this, take a Campbell's soup can, keep it on a rectangular paper and go stretch your imagination. So now we the following options to place the cylinder 1) Base on (6,8). Radius = 3, height = 10. Volume = Pi*R^2*H = 90Pi. 2) Base on (6,10). Radius = 3, height = 8. Dont bother calculating. It is going to be less than 1. 3) Base on (10,8). Radius = 4, height = 6. Volume = 96Pi. Other combinations will either yield a lower volume or be same as 1 and 3. So radius of 4 will bring about max volume for the cylinder.



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Re: retangular box [#permalink]
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22 Feb 2008, 20:56
Many thanks to all of you! OA is B, and I like the shortcut of maratikus. I love gmatclub!
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Re: retangular box [#permalink]
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23 Feb 2008, 10:37
Ah ! Thanks guys for the clear explanation.
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Re: retangular box [#permalink]
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23 Feb 2008, 11:30
rca215 wrote: junkbondswap wrote: Can someone elaborate on this explanation...I dont see how the cyclinder could have a radius greater than 3 if it is to fit into the box
the narrowest point of the box is 6; how could something with a radius of 4 (diameter of 8) fit in there Same here. If 6 becomes the height of the box, then the two remaining side 10, 8 will be l and b and are the only ones that can limit the radius of the cylinder. The minimum of the two which is 8 will become the diameter.



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Re: retangular box [#permalink]
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23 Feb 2008, 12:18
maratikus wrote: sondenso wrote: 191. The inside dimensions of a retangular 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 maximum volume?
A. 3 B.4 C. 5 D. 6. E. 8
Hi friends!, please can help me a shortcut for this? many thanks B is the answer. Here is an easy way to solve it. there are 3 options: the high of the canister is equal to 10,8 or 6. if it's equal to 10 > the radius is min(8,6)/2 = 3 > volume is 10*pi*3^2 = 90*pi if it's 8 > radius is min(10,6)/2 = 3 > volume obviously below the first one if it's 6 > radius is min(10,8)/2 = 4 > volume is 6*pi*4^2 = 96*pi > radius is 4 > B Had jumped to (C) without figuring out the implications. Nicely explained. +1 to poster and maratikus' solution
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Re: retangular box [#permalink]
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23 Feb 2008, 19:26
I will do it this way , in order to get maximum volume of cylinder that can fit in the rectangular box either we maximise the radius or maximise the height
Ist case : maximum height =10 face on that cylinder rest 6X8 , So maximum radius can be 3 volume = pi x 9 x 10=90pi
ii case : maximum radius= can only be 4 as radius of 5 wont fit in as other side is maximum 8 so 10x8 gone to accomodate radius and left 6 for height volume = pi x 16 x 6 = 96pi
so max volume can be with radius 4 Answer is B



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Re: retangular box [#permalink]
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23 Feb 2008, 19:40
neeraj.kaushal wrote: I will do it this way , in order to get maximum volume of cylinder that can fit in the rectangular box either we maximise the radius or maximise the height Neeraj.kausha1 !, your logic is very nice! I love it. Thank you!
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Re: retangular box [#permalink]
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23 Jul 2011, 21:10
B.. Good problem..




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