Differentiation, related rates?

Click For Summary
SUMMARY

The discussion centers on maximizing the area of a rectangle inscribed in a right triangle using calculus. The area \( A \) is expressed as \( A = xH - \frac{H}{b}x^2 \), where \( H \) is the height and \( b \) is the base of the triangle. Participants clarify the differentiation process and suggest using the vertex of the parabola to find the maximum area instead of relying solely on derivatives. The equation of the hypotenuse is given as \( y = H - \frac{H}{b}x \), which is crucial for setting up the problem.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and related rates
  • Familiarity with the properties of right triangles
  • Knowledge of maximizing functions and the vertex of a parabola
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the differentiation of polynomial functions, focusing on the power rule
  • Learn about maximizing area problems in geometry, particularly with inscribed shapes
  • Explore the method of completing the square for quadratic equations
  • Investigate related rates problems in calculus for further applications
USEFUL FOR

Students studying calculus, particularly those focusing on optimization problems, geometry enthusiasts, and educators looking for examples of applying calculus to real-world scenarios.

Willowz
Messages
197
Reaction score
1

Homework Statement


Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown?


Homework Equations


AB/AD = BC/DE


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:
Physics news on Phys.org
Picture added:

lCBl = X

lABl = H

H/h = X/x
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    8.3 KB · Views: 384
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
Replies
11
Views
3K
Replies
3
Views
4K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K