Optimization: A farmer wants to fence an area of 1.5 million square feet in a rectangular field?

SOMEONE ASKED 👇

Optimization: A farmer wants to fence an area of 1.5 million square feet in a rectangular field?

HERE THE ANSWERS 👇

  • For the sake of argument, let’s say that “length” is the dimension that the middle partition is parallel to, and the other two sides are the “width”. Let L be the length and W be the width. Then the area is:

    LW = 1.5 x 10^6 square feet

    The total amount of fencing used, in feet, is going to be the sum of the four sides, plus the partition that runs down the middle:

    (L+L+W+W) + L =

    3L + 2W

    Using the first equation to substitute, you get

    3L + [2(1.5 x 10^6)/L]

    Now that you have the fencing equation in terms of just L, you can take the derivative, set it equal to 0, then solve for L. This tells you the value of L that minimizes the amount of fencing used, which in turn means that it minimizes the cost. To find W, that’s just W = (1.5 x 10^6 square feet)/L.

  • Let the dimensions of the rectangle be x and y, so the length of fencing required is given by:

    L = 3x + 2y

    And the area, A is given by

    A = xy = 1500000 –> y = 1500000/x

    so L = 3x + 3000000/x

    dL/dx = 3 – 3000000/x^2 = 0 (at maximus)

    x^2 = 1000000

    x = 1000 –> y = 1500

    So your answer is 1500 ft, 1000 ft

    Source(s): I know stuff.
  • p = width of the field

    q = length

    pq = 1.5 x 10^6 ft²

    Perimeter to be minimized = 2p + 3q

    P = 2p + 3(1.5 x 10^6 / p)

    P = 2p + (4.5 x 10^6)p^-1

    dP/dp = 2 + (4.5 x 10^6)(-1)p^-2

    dP/dp = 2 – (4.5 x 10^6) / p²

    Set dP/dp = 0 to find the minimum…

    2 – 4.5 x 10^6 / p² = 0

    2p² – 4.5 x 10^6 = 0

    2p² = 4.5 x 10^6

    p² = 9 x 10^6

    p = √(9 x 10^6)

    p = 3000 ft <===================

    q = (1.5 x 10^6) / 3000 = 500 ft <==============

  • enable container have sides L and B and enable fence be parallel to section L Perimeter + divider = 3L + 2B………………..(a million) element of container = L*B = 3000000 So, B = 3000000/L replace this value of B in (a million) Fencing = 3L + (6000000/L) or (3L^2 + 6000000) / L

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