Optimization Problem....?

A garden is to be laid out in a rectangular area and protected by a chicken wire fence. What is the larges possible area of the garden if only 100 running feet of chicken wire is available for the fence?

Thanks for any help!

3 Answers

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  • 1 decade ago
    Favorite Answer

    let length = x

    breadth = 50-x

    area = x*(50-x)

    differntiate it with respect to x and equate it to zero!

    50-2x = 0 that is x=25; 50-x = 25

    so the largest area is 25*25=625 square feet!

  • gp4rts
    Lv 7
    1 decade ago

    The fence forms the perimeter of the garden. The perimeter is P = 2L+2W, where L and W are the length and width of the garden. The area is A =LW. Solve for L or W from the perimeter equation: for example L = .5*(P - 2W); then plug that in for L in the area equation to get an equation in W:

    A = .5*(P- 2W)*W

    A = .5*PW - W^2

    Take the derivative of A to get A' = .5*P - 2W

    set this equal to 0 and sovle for W. You will get w = P/4. From the perimeter equation you will get L = P/4. In other words, the gardent is a square with 1/4 of the fence length forming each side.

  • raj
    Lv 7
    1 decade ago

    material needed=2pir(h+r)

    volume=pir^2h

    pir^2h=1000cc (given)

    h=1000/(pir^2)

    substituing h in the equation 2pir(h+r)

    material=2pir(1000/pir^2+r)

    =2pir*1000/pir^2+2pir^2

    dA/dt=-2000/r^2+4pir

    setting this to zero

    -2000+4pir^3=0

    4pir^3=2000

    r^3=2000/4pi

    r=cube root of 500/3.14

    =5.4 approx

    substituting

    h=1000/(3.14*5.4*5.4)

    =10.9 aapprox

    so the radius is 5.4 cm approx

    and the height 10.9 approx

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