Fence problem?

ok, i need some help with this homework. i need to fence a portion of land so as to provide the greatest area. one side of the land boarders a river so it does not need to be fenced. i have 900 feet of fence to use, how do i do this? i remember that you get some formula and graph it and the vertex is the answer. help please?

5 Answers

  • 1 decade ago
    Favorite Answer

    Let L be the length of the fence parallel to the river and let W be the length of the two fences perpendicular to the river.

    So 2W + L = 900

    Solve for L:

    L = 900 - 2W.

    The area would be W x L.

    Substituting in for L.

    Area = W (900 - 2W)

    = -2W² + 900W

    Now maximize that by taking the derivative and setting it to zero.

    -4W + 900 = 0

    4W = 900

    W = 225

    If you don't know calculus, you could just graph y = -2x² + 900x, for various values of x. You'll see that it is an upside down parabola that is symmetric around the point x = 225 where it has its maximum.

    However you figure it, the area will be maximized when you build a rectangle (half square) with dimensions of 225 feet x 450 feet. Total fencing will be 225 + 450 + 225 or 900 feet, and the area will be 101,250 sq. ft. (Notice how this is more than the intuitive answer of 300 x 300 or 90,000 sq. ft.)

  • Anonymous
    1 decade ago

    If you want to use a graph, you are probably looking for a rectangular area instead of a semi-circle. Let x, x and y be the sides of your area. The area is xy and the length of fence is 2x+y=900. You have two equations in two unknowns. y=900-2x, so the area is x(900-2x). You can graph that and find the maximum or if calculus is allowed, take the derivative and set it to zero.

  • Helmut
    Lv 7
    1 decade ago

    Assuming the river on one side is perfectly straight:

    The maximum area will be enclosed by a semicircle of radius 900/PI, or 128,915.5 square feet.

  • 1 decade ago

    Try 300'x300'x300' otherwise you will have to make a semicircle(pi*r^2) for the most land

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  • 1 decade ago

    bandf beat me to the right answer - good job!

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