Calculus

A rancher wants to construct two identical rectangular corrals using 300 feet of fencing. The Ranger decides to build them adjacent to each other, so they share something on one side. What dimension to the range are used to construct each corral so that together, they will enclose the largest possible area?

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  1. Rancher whave 3 pieces of fencing of the length L ( two on the border of the fence and one in the middle )

    and 2 pieces of fencing of the length W.

    3 L + 2 W = 300

    Divide both sides by 3

    L + 2 W / 3 = 100

    Subtract 2 W / 3 to both sides

    L = 100 - 2 W / 3

    A = L ∙ W = ( 100 - 2 W / 3 ) ∙ W

    A = 100 W - 2 W ² / 3

    Derivative:

    dA / DW = 100 - 2 ∙ 2 W / 3

    dA / DW = 100 - 4 W / 3

    Function have extreme in point where the derivative is zero.

    dA / DW = 100 - 4 W / 3 = 0

    Add 4 W / 3 to both sides

    100 = 4 W / 3

    Multiply both sides by 3

    300 = 4 W

    300 / 4 = W

    75 = W

    W = 75 ft

    Second derivative test:

    If second derivative < 0, then function has a maximum

    If second derivative > 0 then function has a minimum

    In this case secons derivative = d ( 100 - 4 W / 3 ) / dW = - 4 / 3 < 0

    So for W = 75 area has maximum

    L = 100 - 2 W / 3

    L = 100 - 2 ∙ 75 / 3

    L = 100 - 150 / 3

    L = 100 - 50

    L = 50 ft

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  2. Note that after all the math, maximum area happens when the fencing is divided equally among lengths and widths.

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    oobleck

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