BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

volume

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 228
Joined: Sun Aug 17, 2008 8:08 am
Thanked: 4 times

volume

by jainrahul1985 » Wed Jan 05, 2011 10:14 pm
A rectangular box has the dimensions 12 inches x 10 inches x 8 inches. What is the
largest possible volume of a right cylinder that is placed inside the box?

OA 200pi

Please let me know any short cut to slove this question .

Source : Manhattangmat
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Wed Jan 05, 2011 10:28 pm
Solution:
Volume of cylinder is pi*r^2*h.
For maximum volume, radius should be maximum because it has a square power.
This is possible if we take the base of cylinder as the base of the box which is 12*10.
Then the diameter is 10 for the cylinder, as it has to be lesser of 10 and 12.
Or radius is 10/2 = 5 and height is 8.
So volume is pi * 5^2 * 8 = 200pi.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Wed Jan 05, 2011 10:28 pm
jainrahul1985 wrote:A rectangular box has the dimensions 12 inches x 10 inches x 8 inches. What is the
largest possible volume of a right cylinder that is placed inside the box?

OA 200pi

Please let me know any short cut to slove this question .

Source : Manhattangmat
Volume of a Cylinder = pie* radius^2 * height

In the concerned rectangular boxes the
1) base can be 12 * 10 and height 8
or 2) base can be 10*8 and height 12
or 3) base can be 8*12 and height 10

In each case of base the Radius will be the half of the length of width (lesser length)

Case 1. r = 5 h =8 then V = pie(5)^2 (8) = 200pie
Case 2. r = 4 h =12 then V = pie(4)^2 (12) = 182pie
Case 3. r = 4 h =10 then V = pie(4)^2 (10) = 160pie

Answer is 200pie
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Post new topic Post Reply
• Page 1 of 1