The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters....

[VJesus12]

The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

Answer: C

[Brent@GMATPrepNow]

The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

Answer: C

If the OUTER dimensions are 8 cm x 10 cm x 12 cm, and each side is 0.5 cm thick, then the INNER dimensions are 7 cm x 9 cm x 11 cm

Volume of cylinder = pi(radius²)(height)

There are 3 different ways to position the cylinder (with the base on a different side each time).
You can place the flat BASE of the cylinder on the 7x9 side, on the 7x11 side, or on the 9x11 side

If we place the base on the 7x9 side, then the cylinder will have height 11, and the maximum radius of the cylinder will be 3.5 (i.e., diameter of 7).
So, the volume of this cylinder will be (pi)(3.5²)(11), which equals (12.25)(11)(pi), which is a little bit more than 132pi

ASIDE: There's a nice trick for quickly calculating (in your head) the squares of numbers ending with 5 (e.g., 3.5²). See: https://www.gmatprepnow.com/module/gmat ... video/1024

If we place the base on the 7X11 side, then the cylinder will have height 9, and the maximum radius of the cylinder will be 3.5 (i.e., diameter of 7).
So, the volume of this cylinder will be (pi)(3.5²)(9), which equals (12.25)(9)(pi), which is a little bit more than 108pi

If we place the base on the 9x11 side, then the cylinder will have height 7, and the maximum radius of the cylinder will be 4.5 (i.e., diameter of 9).
So, the volume of this cylinder will be (pi)(4.5²)(7), which equals (20.25)(7)(pi), which is a little bit more than 140pi

So, the greatest possible volumeis a little bit more than 140pi, and this occurs when the radius is 4.5


Here's a similar practice question: https://gmatclub.com/forum/the-inside-d ... 28053.html

Answer: C

Cheers,
Brent