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Math Help - 3 Rectangular Gardens

  1. #1
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    3 Rectangular Gardens

    A gardener had 1500 feet of fencing to enclose three adjacent rectangular gardens. Determine the demensions that will produce a maximum enclosed area.
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    Quote Originally Posted by magentarita View Post
    A gardener had 1500 feet of fencing to enclose three adjacent rectangular gardens. Determine the demensions that will produce a maximum enclosed area.
    The area of the 3 gardens is calculated by:

    a = l\cdot w...... (A)

    The fence of all 3 gardens is calculated by:

    f = 2l + 4w = 1500 ...... (B)

    Solve (B) for l. Plug in this term into equation (A):

    a(w)=w\cdot \left( \frac12 f - 2w \right) = -2w^2 + \frac12 f w

    To get the maximum area determine the first derivation of a and solve the equation a'(w) = 0 for w. I've got w = \frac18 f

    That means the maximum area is

    a\left(\frac18 f\right)= -2\left( \frac18 f \right)^2 + \frac12 f \cdot \frac18 f = \frac1{32} f^2
    Attached Thumbnails Attached Thumbnails 3 Rectangular Gardens-drei_acker.png  
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    ok...

    Quote Originally Posted by earboth View Post
    The area of the 3 gardens is calculated by:

    a = l\cdot w...... (A)

    The fence of all 3 gardens is calculated by:

    f = 2l + 4w = 1500 ...... (B)

    Solve (B) for l. Plug in this term into equation (A):

    a(w)=w\cdot \left( \frac12 f - 2w \right) = -2w^2 + \frac12 f w

    To get the maximum area determine the first derivation of a and solve the equation a'(w) = 0 for w. I've got w = \frac18 f

    That means the maximum area is

    a\left(\frac18 f\right)= -2\left( \frac18 f \right)^2 + \frac12 f \cdot \frac18 f = \frac1{32} f^2
    Can this be done without using derivatives?
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  4. #4
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    Quote Originally Posted by magentarita View Post
    Can this be done without using derivatives?
    Of course. The graph of the function a(w) is a parabola opening down. Thus the maximum occurs at the vertex of the parabola:

    a(w)=-2w^2+\frac12fw = -2\left(w^2-\frac14fw{\color{red}{+\left(\frac18f  \right)^2}}  \right){\color{red}{+2\left(\frac18f  \right)^2}}

    The vertex has the coordinates V\left(\frac18f\ ,\ \frac1{32}f^2  \right)

    Therefore the area has a maximum of \frac1{32}f^2 at w = \frac18f
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  5. #5
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    ok....

    Quote Originally Posted by earboth View Post
    Of course. The graph of the function a(w) is a parabola opening down. Thus the maximum occurs at the vertex of the parabola:

    a(w)=-2w^2+\frac12fw = -2\left(w^2-\frac14fw{\color{red}{+\left(\frac18f \right)^2}} \right){\color{red}{+2\left(\frac18f \right)^2}}

    The vertex has the coordinates V\left(\frac18f\ ,\ \frac1{32}f^2 \right)

    Therefore the area has a maximum of \frac1{32}f^2 at w = \frac18f
    I thank you for your help.
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