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Differentiation, related rates?

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown?

    2. Relevant equations
    AB/AD = BC/DE

    3. The attempt at a solution
    I've been trying to do this problem. I looked online and saw an explanation without directly using the area of the triangle, found http://answers.yahoo.com/question/index?qid=20101128161728AA21pl7".

    Can someone remind me how to differentiate A = x H - (H/X)x^2 , in the last part of the problem? Is this from related rates?
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Nov 19, 2011 #2
    Picture added:

    lCBl = X

    lABl = H

    H/h = X/x

    Attached Files:

  4. Nov 19, 2011 #3


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    Science Advisor

    I'm not sure exactly what you want here. You have not given any dimensions for the triangle.

    Assuming a right triangle of base b and height h, we can set up a coordinate system having the two legs as axes. Then the hypotenuse is a line from (0, h) to (b, 0). The equation of that line is y= h- (h/b)x. Let a be the length of the rectangle along the x-axis, b, the length of the rectangle along the y-axis. Since the fourth vertex must lie on the hypotenuse, a and b must satisfy b= h- (h/b)x. Maximize the area, ab, with the constraing b= h- (h/b)a. Yes, that gives A= a(h-(h/b)x)= ah- (h/b)x2, just what you have with a slightly different notation.

    As far as differentiating Hx- (H/X)x2, one of the very first derivative rules you learned is that the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex]. However, you don't really need to use the derivative to maximize this. Complete the square to find the vertex of the parabola.
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