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Calculus

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A fence is 1.5m high and is 1m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Calculate the minimun lenght of the ladder.

I drew a diagram and i ended up with a right angle triangle. i made the hypotenuse the lenght of the ladder, h=1.5 and base=1. I then used the pytagorean theorem to solve for the hypotenuse, but my answer doesnt match the one at the back of the book(4.5m)

  • Calculus -

    Let A be the angle that the ladder makes with the horizontal. The ladder's length L must be such that
    1.5/sinA + 1/cos A = L

    Compute dL/dA and set it equal to zero to get the angle A for which the length is minimum.

    dL/dA = -1.5 cosA/sin^2A + sinA/cos^2A = 0
    1.5 cos^3A = sin^3A
    tan^3A = 1.5
    tanA = 1.145
    A = 48.86 degrees
    L = 1.5/sin48.86 + 1/cos48.86
    = 1.992 + 1.520 = 3.512

    I don't agree with your book's answer, either.

  • Concur with dwrls, book is wrong - Calculus -

    I tried it all algebraically, no trig.

    let the ladder reach y m up the wall, and touch the ground x m from the fence.
    So I had two similar right angled triangles and
    1.5/x = y /(1+x)
    xy = 1.5 + 1.5x
    2xy = 3 + 3x
    y = (3+3x)/(2x)

    L^2 = y^2 + )1+x)^2
    = [(3+3x)/(2x)]^2 + (1+x)^2
    = (9 + 18x + 9x^2)/(4x^2) + 1 + 2x + x^2
    =(9/4)x^-2 + (9/2)x^-1 + 9/4 + 1 + 2z + x^2

    2L(dL/dx) = (-18/4)x^-3 - (9/2)x^-2 + 2 + 2x
    = 0 for a max/min of L

    (-18/4)x^-3 - (9/2)x^-2 + 2 + 2x
    = 0
    times 4x^3
    -18 - 18x + 8x^3 + 8x^4 = 0
    8x^3(1+x) - 18(1+x) = 0
    (1+x)(8x^3 - 18) = 0
    x = -1, silly answer or
    x^3 = 18/8
    x = cuberoot(18)/2 = 1.31037
    sub that back into y = ..
    y = 2.6447

    then L^2 = y^2 + (1+x)^2
    gave me
    L = 3.5117

  • Calculus -

    dwls solution is corret but when you plug 48.86 degrees back into the original equation your calculator must be in rads and you will get 4.5m

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