Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The inside dimensions of a retangular wooden box are 6 [#permalink]
22 Feb 2008, 03:07

1

This post received KUDOS

1

This post was BOOKMARKED

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 _________________

Re: retangular box [#permalink]
22 Feb 2008, 06:10

1

This post received KUDOS

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

Re: retangular box [#permalink]
22 Feb 2008, 07:13

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.

Re: retangular box [#permalink]
22 Feb 2008, 18:05

1

This post received KUDOS

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.

Re: retangular box [#permalink]
23 Feb 2008, 10:30

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.

Re: retangular box [#permalink]
23 Feb 2008, 11:18

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 _________________

Re: retangular box [#permalink]
23 Feb 2008, 18:26

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

Re: retangular box [#permalink]
23 Feb 2008, 18:40

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! _________________

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...