Knowing how it is second nature for my friend Y-san to solve extremal problems by equating derivatives of functions to zero, I was delighted to find a problem that can be solved with only a pencil, some drawing paper and a little thought. Here it is.
Imagine an empty can that weighs 1.5 ounces. It is a perfect cylinder with weightless top and bottom, so that any asymmetry introduced by punching holes in the top can be disregarded. The can holds 12 ounces of beer; so its total weight, when filled, is 13.5 ounces. The can is 8 inches high. Without using calculus, determine the level of the beer at which the center of gravity is at its lowest point.
Consider that, as beer is removed from a full can, the center of gravity is lowered. But empty and full, the center of gravity is in the same place. So there must be a filling factor where it is lowest.
Posted · Report post
Knowing how it is second nature for my friend Y-san to solve extremal problems by equating derivatives of functions to zero, I was delighted to find a problem that can be solved with only a pencil, some drawing paper and a little thought. Here it is.
Imagine an empty can that weighs 1.5 ounces. It is a perfect cylinder with weightless top and bottom, so that any asymmetry introduced by punching holes in the top can be disregarded. The can holds 12 ounces of beer; so its total weight, when filled, is 13.5 ounces. The can is 8 inches high. Without using calculus, determine the level of the beer at which the center of gravity is at its lowest point.
Consider that, as beer is removed from a full can, the center of gravity is lowered. But empty and full, the center of gravity is in the same place. So there must be a filling factor where it is lowest.
Share this post
Link to post
Share on other sites