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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 14, 2012 2:35 PM
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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. Some of the methods I
> tried were;
> a^2 + b^2 = c^2 => m^2x^2 + x^2/m^2 = 4x^2 - 40x +
> 100
> => x^2(m^2+ 1/m^2) = 4x^2 - 40x + 100 => m^2 + 1/m^2
> = (4x^- 40x + 100)/x^2
> Then I took the derivatives of both sides (with
> respect to m on the left side and to x on the right)
> and got:
> 2m - 2(1/m^3) = 40 * (x-5)/x^3 which gave me: x=5 and
> m=1, but since 5 is the root of one of our lines
> (y=10-2x), we can't use x=5 in any further
> calculations...I think. Whatever I try to do and
> however I try to solve this, I always seem to end up
> with a negative number.
> Anyways. I'm not getting any closer to solving this
> at the moment. Any help will be greatly appreciated.

A simpler approach:

If you visualize a family of right triangles whose vertex is at the origin and whose hypotenuse is the line
y = 10 - 2x, then area is minimized when the right triangle is isosceles. So set equal to each other the
lengths of the two legs, each expressed in terms of m
and then solve for m.

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