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wooden box and a cylindrical canister inside with max radius [#permalink]
17 Mar 2010, 17:30

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

83% (01:40) correct
17% (00:58) wrong based on 14 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

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
17 Mar 2010, 19:02

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

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
17 Mar 2010, 19:26

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. _________________

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
18 Mar 2010, 16:04

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!!! _________________

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
20 Aug 2010, 06:22

1

This post received KUDOS

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;

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
21 Aug 2010, 15:17

1

This post received KUDOS

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!!!!!! _________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Re: wooden box and a cylindrical canister inside with max radius [#permalink]
23 Nov 2011, 08:40

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.

gmatclubot

Re: wooden box and a cylindrical canister inside with max radius
[#permalink]
23 Nov 2011, 08:40