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Calculus/Optimization in Calculus
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Question
You are a lifeguard and are standing on the shore of a calm lake. You see a person in need of help who is 480 meters directly down the shore and 220 meters out. You can run at a rate of 8 meters per second and can swim at the rate of 2 meters per second. How far down the shore would you run before entering the water to swim in order to MINIMIZE the time to get to the person in need of help? How long will it take you to get there? (Tenths)
IF this is the case, run along the short until you feel like jumping and don't take time to compute that answer, for by that time, the person may wash even farth away or drown. That's assuming you can swim and the waters not too cold. And do you know mouth to mouth if you need to revive them? But that's reality, so let's get down to the problem. At least this lake is calm, so we don't have to worry about the person moving.
The total distance down the shore is 480 meters, of which you will run 480-x meters. At this point, you (or the lifeguard) will jump in and swim √(x²+220²).
The total time is (480-x)/8 + [√(x²+220²)]/2. This function is t(x), the time it takes to get there. Compute dt/dx, set it equal to 0, and find what the right value for x is.
The answer can be found and rounded to the nearset tenth.
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Answers by Expert:
Scotto
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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.
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