Suppose you are testing a prototype of a new cell phone to find the maximum height (say, in cm) from which you can drop the phone without breaking it. It's a prototype so you only have two phones to test with. Assume that for finding the minimum breakage height, it doesn't matter how many times you drop a phone at a lower height that doesn't break it. Suppose that when deciding the initial height from which to drop the first phone, your goal is to minimize the maximum number of drops necessary. If you had only one phone, you would have to start from height 1 , and keep going up by one cm until it broke. But with two phones you can follow a better strategy, assuming you have a maxiumum height that you care about, call it hmas - you could do something similar to binary search. For instance, you could drop the first phone from a height of hmas/2. If it broke, you would start dropping the second phone at height 1 and move up by one cm each time until it broke. If not you would drop the first phone again from height 3hmax/4; if it broke you would start dropping the second phone at height (hn/2)+1 and move up by one cm each time until it broke, etc. This would require fewer trials in the worst case than first dropping the phone from height 1cm and moving up by one cm per test. Part A: Making the first drop from height hmw/2 is not the best choice. Why not? You should be able to explain this in one or two sentences. Solution: Part B: What is the best height from which to drop the first phone on the first trial? Justify your answer. I am not looking for a formal proof, just sound reasoning. Solution