Science:Math Exam Resources/Courses/MATH102/December 2013/Question C 04
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Question C 04 |
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Shown in the figure below is the view from above of the path taken by a penguin from point A to a feeding area on the shore at point C. The penguin must choose the point B toward which it starts walking. It takes twice as much energy per unit distance for the penguin to walk over land (AB) as to swim through water (BC). The distance AD is 300 m and the distance DC is 400 m. Calculate the value of the distance x (and hence the location of the point B - see figure) that minimizes the energy spent on the entire trip. |
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. |
Hint 1 |
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First determine all the values in the triangle in terms of x. This will be useful later. |
Hint 2 |
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Next, try to figure out an expression for the function you wish to minimize. It might help organize your thoughts to introduce a constant representing how much energy the penguin uses to swim one metre. |
Hint 3 |
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Don't forget to verify this is a minimum either by
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Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution |
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Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. First we compute the lengths denoted in the diagram in terms of x. Since DC = 400, we have that the length of BC is . As AD = 300, we can use the Pythagorean theorem on triangle ADB to see that the length of AB is equal to .
which is equivalent to minimizing since is a positive constant.
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