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Topic: Trigonometric area optimization
Replies: 13   Last Post: Dec 20, 2012 3:29 PM

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Posts: 15
From: California
Registered: 6/29/09
Re: Trigonometric area optimization
Posted: Dec 17, 2012 1:48 AM
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On December 11, 2012, AdamThor wrote:


Earlier today, I took a math exam and there was one problem that I just couldn't solve. The problem is as follows:

The lines y=10-2x, y=mx and y=(-1/m)x, where m > 1/2, form a right triangle. Find an m, so that the area measurement of the triangle is as small as possible.

I just couldn't, for the life of me, find a suitable function which I could take the derivative of. I made some honest attempts at solving this, but nothing gave me a definitive answer............
A response to the responders:

Gee, I wouldn't have guessed that this topic could run off in a direction of "who solved first" or who deserves
"credit/applause." I have little interest in those aspects, but I will toss in my 2 cents...take it for what it's worth.

I, too, appreciate the OP bringing an interesting problem to our attention and I think that we have all been trying to be helpful. My approach to the query by AdamThor was simply this: he indicated that this came up in an exam and he obviously had tried hard to find a solution, but was unsuccessful, and he wanted some answer. This, to me, shows that the exam had achieved one of its important purposes: help point out what you *don't* know...and possibly entice the testee to study further and fill in weak points in the subject matter (Sometimes, in tests I have designed, I followed the maxim that challenge can be good thing: if you achieve a perfect score on tests, then you haven't really been "tested"!). As an instructor I would bend over backwards to help any student who showed that interest...particularly post-exam.

My attention was drawn to the OP's mentioning his exam but didn't go into details. But since this was posted on a board, we might be able to guess at the appropriate level to address a response. I like to work out problems with trig and calculus as much as anyone, and the OP was obviously already familiar with derivatives. But...we don't know exactly what kind of exam it was, only that this was "one problem (he) just couldn't solve," but I would guess that it was a timed event (versus a "take home" or "take all the time you want"). It might have been "multiple choice" (unlikely) or a "show all your work." (The "multiple choice" becomes ineffective if the testee can just plug in the choices and see which one works, except if one of the choices is "none of the above".) But if this exam was timed then time management can be an important consideration.

When time to solve becomes paramount, it can be useful to assume that there IS a solution that the exam constructor had in mind and that it is achievable within the time constraints. So, this becomes a search for one's mathematical insight or at least for an "Aha!" moment. Most any approach that requires deep calculation or solutions to equations that seem intractable, is not likely to be successful (nor were those the exam's intent). So, even if this particular exam came at the end of a teaching unit on "the use of derivatives in analytical geometry," sometimes an exam question is intended to show you when a specific approach is NOT the most expeditious path.

What was behind my own insight into this problem? I once observed a wire coat-hanger swinging on a nail on a wall. Its shape, other than the hook, was similar to an isosceles triangle (although not a right triangle), and it happened to be hanging not from its hook but rather from the top of its triangular shape. I noticed a horizontal trim line on the wall behind the hanger (but above and parallel to the hanger's bottom when the hanger was at rest). When the hanger would swing back-and-forth in pendulum fashion, the area outlined by the triangle sides and the trim line varied from a larger to a smaller amount. But this was truly an example of triangles of equal height and third sides having a common direction: obviously the areas subtended depended only upon the length of wall trim subtended, and this led to the insight that isosceles yielded the smallest area. It didn't matter whether or not this was proven back in high school geometry class...the common-place demonstration stuck with me!

Thank you AdamThor and all responders for taking the
time/effort and for taking me down a path of questioning methods and motives.

Well...maybe all that was worth 3 cents!

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