bobmck
Posts:
15
From:
California
Registered:
6/29/09


Re: Trigonometric area optimization
Posted:
Dec 14, 2012 2:35 PM


>On December 11, 2012, AdamThor wrote:
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.
************* 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.

