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

Redeem

Vol of cylinder placed in a rectangular box

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 160
Joined: Mon Apr 05, 2010 4:41 am
Thanked: 7 times

Vol of cylinder placed in a rectangular box

by gmat1011 » Sun Jun 20, 2010 4:17 am
Lets say the dimensions of a rectangular box are 20, 30, 40. What is maximum possible volume of a cylinder which can be placed in the box? There is a similar problem in GMAT Guide Vol 5 (Geometry)

I am looking for the concept here... I am aware that volume of a cylinder is (pie) * r^2 * h

I think it boils down to what you take as r for the cylinder... but I am not comfortable with the reasoning for arriving at the r...

Thanks!
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pm
Thanked: 26 times
Followed by:2 members
GMAT Score:760

by mj78ind » Sun Jun 20, 2010 6:40 am
Let us look at the base of the cuboid and fix the base of the cylinder on it. What are the options:

1. the base is 20X30 ........ hence we can have a cylinder of d = 20 that would fit in. The height of the cuboid and the cylinder = 40. the volume = pie*(d/2)^2*h = pie*(20/2)^2*40 = pie*4000

2. 30X40 base. Volume = pie*(30/2)^2*20 = pie*225*20 = pie*4500

3. 20X40 base is not useful since it will always be beat by 20X30 base (diameters remain 20 but the former has height 30, latter has 40).

Hence, the max volume of the cylinder can be 4500*pie with diameter = 30 and height = 20
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Sun Jun 20, 2010 8:01 am
gmat1011 wrote:Lets say the dimensions of a rectangular box are 20, 30, 40. What is maximum possible volume of a cylinder which can be placed in the box? There is a similar problem in GMAT Guide Vol 5 (Geometry)

I am looking for the concept here... I am aware that volume of a cylinder is (pie) * r^2 * h

I think it boils down to what you take as r for the cylinder... but I am not comfortable with the reasoning for arriving at the r...

Thanks!
Let's look at the formula:

volume = pi(r^2)h

So, the two variables are the radius and the height.

Since the formula contains r^2 (and just h^1), the radius is going to have a much bigger impact on the volume than will the height. Accordingly, our goal is to maximize the radius and let the height take care of itself.

So, with dimensions of 20*30*40, what's going to afford the biggest radius? Setting the base on the 30*40 side, so we can have a radius of 30.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 160
Joined: Mon Apr 05, 2010 4:41 am
Thanked: 7 times

by gmat1011 » Sun Jun 20, 2010 8:11 am
Thanks.

Let H = 30

Base = 20*40 (with 40 being the diameter for the base of the cylinder)

Then Volume = pie * 20*20 * 30 = 12000pie

Why cant we do it this way? Maybe this is a silly questions but is it because the circular base of the cylinder wont fit that way in the rectangular bottom of the box? Mathematically how do you make that determination (of selecting the shorter of the 2 sides other than H as the diameter)? It seems intuitive if I try to draw a rough sketch... But want to u/stand this... Thanks.
Top
Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pm
Thanked: 26 times
Followed by:2 members
GMAT Score:760

by mj78ind » Sun Jun 20, 2010 8:19 am
gmat1011 wrote:Thanks.

Let H = 30

Base = 20*40 (with 40 being the diameter for the base of the cylinder)

Then Volume = pie * 20*20 * 30 = 12000pie

Why cant we do it this way? Maybe this is a silly questions but is it because the circular base of the cylinder wont fit that way in the rectangular bottom of the box? Mathematically how do you make that determination (of selecting the shorter of the 2 sides other than H as the diameter)? It seems intuitive if I try to draw a rough sketch... But want to u/stand this... Thanks.
I guess mapping it out is the best way else we will have to get into inequalities which will be pretty ugly compared to the rough sketch :)
Top
Master | Next Rank: 500 Posts
Posts: 160
Joined: Mon Apr 05, 2010 4:41 am
Thanked: 7 times

by gmat1011 » Sun Jun 20, 2010 8:32 am
Many thanks... if there is a way to explain this without inequalities..... thx
Top
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

by Testluv » Mon Jun 21, 2010 2:26 am
gmat1011 wrote:Many thanks... if there is a way to explain this without inequalities..... thx
Mapping it out/visualizing it, and understanding the dimensions involved is the best way here. If we made the diameter 40, then on the other sides, the cylinder would come out of the box. So, we choose the intermediate value for diameter, and confirm that the cylinder can be contained within the box.
Kaplan Teacher in Toronto
Top
Post new topic Post Reply
• Page 1 of 1