Vol of cylinder placed in a rectangular box

[gmat1011]

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!

[mj78ind]

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

[Stuart@KaplanGMAT]

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.

[gmat1011]

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.

[mj78ind]

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

[gmat1011]

Many thanks... if there is a way to explain this without inequalities..... thx

[Testluv]

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.