AdamThor
Posts:
3
From:
Iceland
Registered:
12/11/12


Trigonometric area optimization
Posted:
Dec 11, 2012 10:07 AM


Hello.
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=102x, 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 * (x5)/x^3 which gave me: x=5 and m=1, but since 5 is the root of one of our lines (y=102x), 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.

