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  1. #1
    Dewitt
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    In need of help.

    Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers

    Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.

    1.) Express the area A(x), of the rectangle as a function of X.

    2.) Determine the domain of A(x).

    3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

    This is the problem I am having alot of trouble with as I have no Idea where to even start. Any help anyone could bring to this equation for me would be greatly appreciated, and thnx
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Dewitt View Post
    Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers

    Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.
    We'll call the longer side l

    So: 2x+2l=32

    Then: 2(x+l)=32

    Therefore: x+l=16

    We know that x is less than l, which means that: x+x<x+l

    Substitute and add: 2x<16

    Finally: x<8

    1.) Express the area A(x), of the rectangle as a function of X.
    You know that: A=xl

    Substitute: A=x(16-x)

    And there it is.


    2.) Determine the domain of A(x).
    3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.
    Well x has to be greater than 0, and that x is less than 8.

    Therefore: 0(16-0)<A<8(16-8)

    Subtract: 0(16)<A<8(8)

    Multiply: 0<A<64

    So A(x) is between zero and sixty-four.
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  3. #3
    MHF Contributor
    Joined
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    Quote Originally Posted by Dewitt View Post
    Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers

    Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.

    1.) Express the area A(x), of the rectangle as a function of X.

    2.) Determine the domain of A(x).

    3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

    This is the problem I am having alot of trouble with as I have no Idea where to even start. Any help anyone could bring to this equation for me would be greatly appreciated, and thnx
    Here is one way.

    Pre-Calculus. No differentiation/derivatives yet? Then we use parabolas for maximum A(x).

    A rectangle is a quadrilateral of whose opposite sides are equal in lengths and whose 4 interior angles are all right angles.

    1.) Express the area A(x), of the rectangle as a function of X.

    Since the shorter side is x, and the perimeter is 32 inches, then the longer side is
    (32 -2x)/2 = (16-x) in.

    A(x) = x(16 -x) = 16x -x^2 ------------answer.

    ---------------------------------------------------
    2.) Determine the domain of A(x).

    A(x) = x(16 -x)
    The domain of A(x) is from x=0 up to x=16, or [0,16]. -----------answer.

    ------------------------------------------------------
    3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

    A(x) = 16x -x^2 is a vertical parabola that opens downward---because the coefficient of the x^2 is negative.
    The maximum value is the highest point or vertex of the parabola.

    Let us examine the graph of A(x) = 16x -x^2
    A vertical parabola whose vertex is at (h,k), that opens downward, is
    (y-k) = -a(x-h)
    So,
    Let y = A(x)
    y = 16x -x^2
    y = -(x^2 -16x)
    y = -(x^2 -16x +64 -64)
    y = -(x^2 -16x +64) -(-64)
    y -64 = -1(x -8)^2 ------------------the parabola of A(x).
    Hence, point (8,64) is the vertex.
    Meaning, when x=8 in., the area of the rectangle is maximum and it is 64 sq.in.

    Therefore, maximum area of the rectangle is 64 sq.in. ---------answer.
    when x=8 in.

    The dimensions of the rectangle when maximum:
    x = 8 in.
    (16 -x) = (16 -8) = 8 in. also
    Therefore, when area is maximum, the rectangle is a square of 8 inches per side. -----------anwser.
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