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Maximizing a rectangle with a semicircle on top

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Heres a picture that might help
    http://img132.imageshack.us/img132/3809/semirectanglexf6.png

    A family wants to create that shape for basketball with duct tape, the family only has 20 ft of duct tape. The family decides they want to maximize the area of the rectangle. What dimensions would give the rectangle it's maximum area?

    2. Relevant equations
    H=height of rectangle
    X=diameter of the semicircle,also width of the rectangle
    Perimeteter=20
    20=2H+2X+(1/2)(X*PI)

    3. The attempt at a solution
    Anything to do from here I dont get. I kind of understand how to maximize the whole thing, but just the rectangle confuses me.

    Hope you can help!
     
    Last edited: Dec 7, 2007
  2. jcsd
  3. Dec 7, 2007 #2

    rock.freak667

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    Homework Helper

    Make an expression for the area...and then see what you get
     
  4. Dec 7, 2007 #3
    Alright the area would be
    A=(H*X)+(PI*x/2^2)=(H*X)+(PI*X^2/4)

    I was thinking of substituting H for
    20=2H+2X+1/2X*PI
    H=(20-2x-1/2X*PI)/2

    So
    A=X(20-2x-1/2X*PI)/2
    But using that won't I only be able to find the dimensions that would give the greatest area of the whole thing?

    I understand I need to make the H*X as big as possible, but don't know how.
    This is like my worst subject.
     
  5. Dec 7, 2007 #4

    rock.freak667

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    Homework Helper

    Find [itex]\frac{dA}{dx}[/itex] and at a stationary value, [itex]\frac{dA}{dx}=0[/itex]

    when you find the value for X, find [tex]\frac{d^2A}{dx^2}[/tex] at the value for X, if it is negative then it is a maximum value, so for that value of X, the area will be a maximum.
     
    Last edited: Dec 7, 2007
  6. Dec 7, 2007 #5
    Umm after you wrote find it shows an x, like when pictures dont load, could youo write it out?
     
  7. Dec 8, 2007 #6
    Ok nevermind, but I dont understand what you mean, how do i do that?
     
  8. Dec 8, 2007 #7
    I didn't check the algebra but the method seems right. You have a function for the area with respect to width, so you can find where the width is maximized. You need to graph that and find the abs. max of the function (needs to be in the first quadrant)

    Since you put this in the pre-calculus forum I'm assuming you don't know how to find the area with out a calculator so I wouldn't look at rock's post unless you do know calculus.
     
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