The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics »

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 3
From: Iceland
Registered: 12/11/12
Trigonometric area optimization
Posted: Dec 11, 2012 10:07 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


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.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.