Help Solve Optimization Problem

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SUMMARY

This discussion focuses on solving an optimization problem involving the area of a parabolic segment and a triangle defined by the equation y=x^2. Participants calculated the area of the segment using definite integrals, specifically the formula I=\int_a^b|f(x)-g(x)|\,d\,x, where f(x)=2x+3 and g(x)=x^2. The area of the triangle was determined to be 8 square units, which is 3/4 of the area of the parabolic segment calculated as 32/3 square units. The discussion also clarified the geometric interpretation of definite integrals and the method for finding the maximum distance from a line to a point on the parabola.

PREREQUISITES
  • Understanding of definite integrals and their geometric interpretation
  • Familiarity with the concept of area between curves
  • Basic knowledge of quadratic functions and their properties
  • Ability to apply optimization techniques in calculus
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  • Study the application of definite integrals in calculating areas between curves
  • Learn how to derive equations of lines given two points
  • Explore optimization techniques in calculus, specifically for geometric problems
  • Review the properties of parabolic functions and their intersections with linear functions
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Students studying calculus, particularly those focusing on integration and optimization problems, as well as educators looking for practical examples of applying definite integrals in geometric contexts.

  • #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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