# Differentiation, related rates?

1. Nov 19, 2011

### Willowz

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

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. Nov 19, 2011

### Willowz

lCBl = X

lABl = H

H/h = X/x

#### Attached Files:

• ###### Untitled.jpg
File size:
8.3 KB
Views:
50
3. Nov 19, 2011

### HallsofIvy

Staff Emeritus
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 $x^n$ is $nx^{n-1}$. However, you don't really need to use the derivative to maximize this. Complete the square to find the vertex of the parabola.