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Math: Calculus

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A fence is to be built to enclose a rectangular area of 230 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.

  • Math: Calculus -

    My dimensions that I got is sqrt(215.625)*230/sqrt(215.625) but it's wrong.

  • Math: Calculus -

    cost = 5(2b+L) + 15 L = 10 b + 20 L
    bL = 230 so b = 230/L

    cost = 2300/L + 20 L

    d cost/dL = -2300/L^2 + 20 L

    = 0 for min
    20 L = 2300/L^2
    L ^3 = 115
    L = 4.86
    b = 47.3
    check my arithmetic !!!

  • error -

    cost = 5(2b+L) + 15 L = 10 b + 20 L
    bL = 230 so b = 230/L

    cost = 2300/L + 20 L

    d cost/dL = -2300/L^2 + 20 *******

    = 0 for min
    20 = 2300/L^2
    L ^2 = 115
    L = 10.72
    b = 21.5
    b = 47.3

  • Math: Calculus -

    cost=5*(2w+L)+15L

    Area= Lw or w= ARea/L=230/L

    cost= 10(230/L)+5L+15L

    dcost/dL= -2300/L^2+20=0
    L=sqrt 115 and w= 230/sqrt115

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