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Optimization problem

  1. Jan 22, 2008 #1
    I'm having a tough time solving this question.:frown: I'd appreciate it if someone can please help me out

    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    y=x^2

    Triangle QPR =3/4 of the area of the parabolic segment enclosed between QR and the parabola

    3. The attempt at a solution

    I don't know where to start :(
     
  2. jcsd
  3. Jan 22, 2008 #2

    NateTG

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    Science Advisor
    Homework Helper

    Well, the area of the segment should be pretty easy to work out. Why don't you start there? It might give you some ideas on how to calculate the area of the triangle.
     
  4. Jan 22, 2008 #3
    How do I calculate the area of the segment?
     
  5. Jan 22, 2008 #4
    Do you know how to calculate the area between two functions [tex]f(x),g(x)[/tex] for [tex] x \in (a,b)[/tex]?
     
  6. Jan 22, 2008 #5
    Nope, we haven't been taught that. We actually just started with integration today. Anyways, I think I kind of got the first step though -


    y=x^2

    and then the equation of the line b/w Q and R can be found out
    m = 9-1/ 3-(-1) = 8/4= 2
    y-1=2(x+1)
    y=2x+3

    elminating y and we get:
    x^2 - 2x -3

    now what shall I do?
     
  7. Jan 22, 2008 #6
    The area between two functions [tex]f(x),g(x)[/tex] for [tex] x \in (a,b)[/tex] is

    [tex]\int_a^b|f(x)-g(x)|\,d\,x[/tex]

    In your case [tex]f(x)=2x+3,\,g(x)=x^2[/tex].

    Now can you caclulate the area of the segment?
     
  8. Jan 22, 2008 #7
    We just started integrating today, So I am not that confident about it :( I'll give it a shot though

    2x+3-x^2
    [(2x^2)/2] +3x- [(x^3)/3]
    [x^2]+3x-[(x^3)/3]

    I don't know what to do now
     
  9. Jan 22, 2008 #8
    Correct! But it is a definite integral, so

    [tex]I=\int_a^b|f(x)-g(x)|\,d\,x=(x^2+3x-\frac{x^3}{3})\Big|_{-1}^{3}[/tex]
     
  10. Jan 22, 2008 #9
    :cool:

    now how shall I proceed?
     
  11. Jan 22, 2008 #10
    Did you find I?
     
  12. Jan 22, 2008 #11
    I don't know how to do that. Do I have to plug -1 and 3 into x?
     
  13. Jan 22, 2008 #12
    Ok!
    [tex](x^2+3x-\frac{x^3}{3})\Big|_{-1}^{3}[/tex] means

    [tex](x^2+3x-\frac{x^3}{3})\Big|_{x=3}-(x^2+3x-\frac{x^3}{3})\Big|_{x=-1}[/tex]
     
  14. Jan 22, 2008 #13
    So that just becomes

    9 -(2 - (-1/3))
    9 -(2.33)
    6.67 units^2

    ?
     
  15. Jan 22, 2008 #14
    I think its better [tex]\frac{32}{3}[/tex] :smile:
     
  16. Jan 22, 2008 #15
    Ooops! lemme recheck
    we first plug in 3 and get

    [9+9 - (27/3)]
    =9

    and for the other side we get

    -[1-3 - [-1/3] ]
    -[-2 + [1/3]]
    -[-1.67]
    1.67

    9+1.67 = 10.67 or 32/3 :)

    aah so the area of the parabolic segment = 32/3 units^2
    while the area of the Triangle = (3/4)*(32/3) = 96/12 = 8 units^2
     
  17. Jan 22, 2008 #16
    Yes! But I think you have to calulate the area of the triangle independently.
     
  18. Jan 22, 2008 #17
    Thank you! :smile:

    But doesn't it say that "Triangle QPR =3/4 of the area of the parabolic segment enclosed between QR and the parabola"

    so that makes the triangle areas 3/4 of 32/3units^2 which is equal to 8..right?

    also can you please explain how you got 3 and -1 for the numbers to be plugged in?


    Thanks
    Jennie
     
  19. Jan 22, 2008 #18
    The geometric meaning of the definite integral

    [tex]\int_a^b|f(x)| \,d\,x [/tex]

    is the area between the graph of [tex]f(x)[/tex] the x-axis and the vertical lines [tex]x=a,x=b[/tex]

    When you have two curves and you want the area between them you should use

    [tex]\int_a^b|f(x)-g(x)| \,d\,x [/tex]

    where now [tex]x=a[/tex] and [tex]x=b[/tex] are the points where the two urves intersect. In your case these are [tex]x=-1, \,x=3[/tex].

    From the statement of the problem I think that first you should calculate the area of the triangle too, in order to see that Arcimedes was correct! :smile:

    But I don't I take an oath! :rofl:
     
  20. Jan 22, 2008 #19
    Thanks for the explanation about the definite integral :smile:

    Now how do we go about calculating the area of the triangle?
     
  21. Jan 22, 2008 #20
    I don't understand what [tex]\overline{QR}[/tex] stands for. Does it stands for the chord QR?
     
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