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Math Help - PreCal Rectangle

  1. #1
    Newbie zacharyrod's Avatar
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    Exclamation PreCal Rectangle

    Alright guys... So I was given a problem on a test for extra credit, but couldn't solve it. Regardless, I won't let it go. So here it is:
    You are given a narrow piece of wood that is 90cm long. You want to split the 90cm of wood into 4 pieces so as to make a rectangular frame.
    Let x be the length of a short side of the rectangle. Then x is also the length of a short piece of wood.
    Write an expression for the length of a long side, then a function of x that gives the rectangles area.

    Now I got this: l= 45-x, but I'm pretty sure it is wrong because it doesn't fit into the function for the area. I think a formula for maximum volume would help, but I can't remember it.

    Anyway, I would really appreciate it if someone could show me how to find the long side. Thanks!
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by zacharyrod View Post
    Alright guys... So I was given a problem on a test for extra credit, but couldn't solve it. Regardless, I won't let it go. So here it is:
    You are given a narrow piece of wood that is 90cm long. You want to split the 90cm of wood into 4 pieces so as to make a rectangular frame.
    Let x be the length of a short side of the rectangle. Then x is also the length of a short piece of wood.
    Write an expression for the length of a long side, then a function of x that gives the rectangles area.

    Now I got this: l= 45-x, but I'm pretty sure it is wrong <<<<<<< that's wrong: Your equation is OK
    because it doesn't fit into the function for the area. I think a formula for maximum volume would help, but I can't remember it.


    Anyway, I would really appreciate it if someone could show me how to find the long side. Thanks!
    The area of a rectangle is calculated by:

    A = length \cdot width

    You got: w = x and l = 45 - x. Therefore

    A(x)=x(45-x)~\implies~A(x)=45x-x^2

    That means:
    • A is a parabola opening down with the maximum at it's vertex

    • The maximum occurs if A'(x) = 0


    I don't know which method you prefer to answer the question.
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  3. #3
    Newbie zacharyrod's Avatar
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    Seoul, Korea
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    Red face

    Quote Originally Posted by earboth View Post
    The area of a rectangle is calculated by:

    A = length \cdot width

    You got: w = x and l = 45 - x. Therefore

    A(x)=x(45-x)~\implies~A(x)=45x-x^2


    That means:
    • A is a parabola opening down with the maximum at it's vertex

    • The maximum occurs if A'(x) = 0

    I don't know which method you prefer to answer the question.
    Thanks for helping me realize that!
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