Max/Min Problem: Find Max Area of Rectangle Bounded by X-Axis & Semicircle

  • Thread starter Thread starter Riogho
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves finding the dimensions of a rectangle that is bounded by the X-axis and a semicircle defined by the equation \( y = \sqrt{36 - x^2} \). The goal is to determine the dimensions that maximize the area of the rectangle.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the rectangle's width and height, with one noting that the width is \( 2x \) and height is \( y = \sqrt{36 - x^2} \). Questions arise about how to express the area as a function of \( x \) and how to find its maximum.

Discussion Status

Participants are engaged in exploring the formulation of the area function and the process of finding its maximum. There is a progression from defining the area to discussing the differentiation needed to find critical points, indicating a collaborative exploration of the problem.

Contextual Notes

There is an emphasis on the understanding of basic geometric principles, such as the area of a rectangle, and the discussion includes some clarification on terminology used in the context of the problem.

Riogho
Messages
119
Reaction score
0

Homework Statement


A rectangle is bounded by the X-axis and the semicircle Y = [(sqrt)36-x^2]. What dimensions should the rectangle have so that its area is a maximum.

Homework Equations



Just a note, the 36-x^2 is all under the radical.

The Attempt at a Solution



Ditto.
 
Physics news on Phys.org
Well, the width of the rectangle is clearly [itex]2x[/itex], and its height is [itex]y=\sqrt{36-x^2}[/itex]...so what is its area [itex]A(x)=?[/itex] for a given value of [itex]x[/itex]? How do you find the maximum of such a function?

P.S. I don't think the definition of "ditto" is exactly what you seem to think it is (just a friendly FYI)
 
Wouldn't I just find the area by multiplying both of them together?
 
Sounds good to me; the last time I checked the area of a rectangle was just width times height ;0)
 
So I multiply them together then take the derivative, and then find the critical points of that?
 
Yes, now do it!
 
Done.

Many thanks.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 36 ·
2
Replies
36
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K