Help Solve Optimization Problem

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Homework Help Overview

The discussion revolves around an optimization problem involving the area of a triangle and a parabolic segment defined by the equation y=x^2. Participants are exploring how to calculate the area of the triangle QPR, which is stated to be 3/4 of the area of the parabolic segment enclosed between points QR and the parabola.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the area of the parabolic segment and the triangle, questioning how to find the area between two functions and the specific coordinates involved. There is also exploration of the distance from a point to a line and its implications for the triangle's area.

Discussion Status

The discussion has progressed through various attempts to calculate areas and clarify concepts related to definite integrals and geometric interpretations. Some participants have provided guidance on how to approach the calculations, while others are still seeking clarification on specific steps and concepts.

Contextual Notes

Participants note that they have just begun learning about integration, which may limit their confidence in applying these concepts. There is also mention of the need to calculate the area of the triangle independently, as well as the importance of understanding the coordinates of the vertices involved.

  • #31
Rainbow Child said:
Good! :smile:

Now what's the maximum value of d? For which x_0 you obtain that?

:biggrin:

The max value I get is when x_0=1 and D would then be 4/sqrt(5)
 
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  • #32
Very nice! Let us now be proud of ourselves and calculate the area of the triangle! :smile:
 
  • #33
Rainbow Child said:
Very nice! Let us now be proud of ourselves and calculate the area of the triangle! :smile:


So area of the triangle = 1/2 * 8.944 *4 = 17.88 units^2

I got the midpoint of QR as (1,5) and the distance from there was 4.

So my final answer for the area of the triangle is 17.88 units^2

:-p
 
  • #34
No, the distance is d=\frac{4}{\sqrt{5}}=height. You don't need the midpoint
 
  • #35
ohhh. then the Area = 1/2 * 8.944 * 4/sqrt(5) = 7.999 which is = 8units^s [our earlier answer :smile:]
 
  • #36
Bravo! :smile:

Archimedes was right! :smile:
 
  • #37
cool! :cool: Thanks for all the help Rainbow Child!

really appreciate it! :-p
 

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