Geometry Question

nadib002 wrote:The rectangular box has the dimensions 12 inches * 10 inches * 8 inches. What is the largest possible volume of a right cylinder that is placed inside the box?

The answer says that -- "The radius of the cylinder must be equal to half of the smaller of the 2 dimensions"-- I cannot seem to comprehend the statement. Any help would be appreciated.


Thank you

Imagine a rectangle, and then imagine about the largest possible circle that can be drawn inside of it. What could be the maximum diameter of the circle? Is it equal to the longer side of the rectangle? No, else it won’t anymore be inside the rectangle. So, is it equal to the shorter side of the rectangle? Of course yes. Hence, if the diameter must be equal the smaller of the two dimensions, then the radius must be equal to half of the smaller of the two dimensions. Nothing is wrong in there.

Volume of the cylinder is TTr²h , keep aside TT which is a constant.
The dimensions of the rectangle are 12,10 and 8
Three cylinders are possible, each with a diff. height 12, 10 or 8.
Consider the cylinder with height 8
Now the diameter cannot be 12 it has to be 10
Similarly if the height is 12
the diameter cannot be 10 but lesser of the two i.e 8.

In a rectangle of dimension 12 and 10 , what is the diameter of the largest circle possible inside the rectangle ?
Will a circle of radius 6 fit inside the rectangle ?

The trick is to find the 3 different combination:

1. Base with 12 & 10, and height 8.
2. Base : 10 & 8, Height 12.
3. Base: 12 & 8, height.

The smaller side of the base is diameter.
Calculate ares of the cylinder in all 3 scenarios:

1. pi* 5^2*8 = 200pi.
2. pi* 4^2 * 12 = 192pi.
3. pi* 4^2 * 10 = 160pi.

Hence the answer is 200pi.