Question is:

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?

I don't get that here is only three cases. is the keyword here "right cylinder" that the height has to be the 12(largest number), and the other two lower numbers are the bases(diameters used to determine the radius).?

Why cannot it be Case 4, where r=5 and h=12? Which yields 300 pie, a higher number.

I'm looking for more clarity as to some of the math rules being applied here that I am not aware of. Like the height is the remaining dimension of the box, the 3 cases doesn't cover all the bases to me at least.

It might help you to actually draw the cases on paper. One way is to draw the three cases for the box "bottom": 8 by 10 rectangle, 8 by 12 rectangle, and 10 by 12 rectangle. Then just draw a circle to represent the cylinder "down" view --imagine yourself looking into the box at the can you have placed there.


If your cylinder is 12 tall, it would stick out of any box that is only 8 or 10 high. Our box must be sitting with the 12 edge vertical, and your 12 inch tall cylinder is therefore sitting on the 8 by 10 box bottom.

So if your cylinder is sitting on the 8 by 10 bottom of the box, how large can the diameter be? The diameter can be at most 8--if you draw it you will see that such a cylinder is tangent to the long sides of the 8 by 10 rectangle. The diameter CANNOT be 10--draw the circle and you will see that it juts outside of the rectangle.

So while your Case 4 is a cylinder with larger volume, it doesn't fit in the box, a requirement of the problem.

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Emily Sledge
Instructor
ManhattanGMAT