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 500,000 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:

(If h=12, r can't be more than 4. So pi*r^2*h=12*16pi=192pi
Apparently h=10, r=4 can't be more

Since vol. involves r^2, looking for a greater r would help. However r can't be 6 because the max possible length of the other side could be only 10 which could not fit a circular base)

Remember: the base of the cylinder is a circle
r is constant around the circle, so the base can completely fit in a square, not a rectangle. Therefore, you will not be able to use the entire space in the base of the rectangular box.

the volume of a cylinder is pi * r^2 * h

select largest possible r such that the dimensions of the rectangular base allow maximum value or r
you would want to maximize r more than h because r is squared here
so, select bases 10 and 12 allowing a maximum value of r = 5

volume of cylinder: 25 pi * 8 = 200 pi

to check, think if 8 is one of the dimensions of the base, r = 4
volume of cyclinder: 16 pi * (either 10 or 12 ) = 160 pi or 192 pi
not maximum volume

Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

Re: A rectangular box has dimensions 12*10*8 inches. What is the [#permalink]

Show Tags

18 Jul 2013, 02:15

4

This post received KUDOS

Since the radius of the cylinder is squared to obtain its volume, the radius should be the maximum possible radius in order for the cylinder to achieve its maximum volume. Point to keep in mind while selecting the maximum possible radius: The diameter should be <= two of the largest sides of the rectangle. If one selects the largest side of the rectangle as the diameter, then the cylinder won't fit into the other side of the rectangle. Refer Figure.

Attachment:

cylinder in rectangle.JPG [ 10.15 KiB | Viewed 9759 times ]

As a rule of thumb, in such problems, select the second largest side as the diameter (note that it is the diameter and one has to calculate the radius by dividing by 2 before calculating the volume). And the left alone smallest side will be the height of the cylinder (as you need the two largest sides to enclose the bottom of the cylinder the only choice left out for height is the smallest side).

Last edited by Jaisri on 18 Jul 2013, 02:19, edited 1 time in total.

Re: A rectangular box has dimensions 12*10*8 inches. What is the [#permalink]

Show Tags

08 Oct 2013, 06:15

willget800 wrote:

A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie B. 200 Pie C. 300 Pie D. 320 Pie E. 450 Pie

Always remember, the largest possible value will have a diameter that will need to feet on 2 of the sides. Therefore, if it fits in 10 it will also fit in 12. So choose a diameter of 10 = 2r, r=5 and then use the other side 8 for the height giving a total volume of 200pi

Re: A rectangular box has dimensions 12*10*8 inches. What is the [#permalink]

Show Tags

27 Dec 2013, 15:52

willget800 wrote:

A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie B. 200 Pie C. 300 Pie D. 320 Pie E. 450 Pie

So first we need to place the cylinder so that its diameter is >= to at least two of the dimensions. That is 10 and 12, so r = 5

Next, height has to be 8 then. I suggest you draw the cuboid and see for yourself that we need a side where the circumference stands to be 10*12 otherwise diameter won't fit

Then I guess the question asks for the largest volume possible

Re: A rectangular box has dimensions 12*10*8 inches. What is the [#permalink]

Show Tags

08 Aug 2015, 04:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A rectangular box has dimensions 12*10*8 inches. What is the [#permalink]

Show Tags

14 Oct 2015, 02:23

Bunuel wrote:

Ashamock wrote:

Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...