Dimensions...Calculus 1?

The sum of the perimeters pf an equilateral triangle and a square is 20. Find the dimensions of the triangle and the square that produce a minimum total area.

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  • Best Answer

    3t + 4s = 20

    A = (1/2) * t * sqrt(t^2 - (t/2)^2) + s^2

    A = (1/2) * t * t * sqrt(1 - 1/4) + s^2

    A = (1/2) * t^2 * sqrt(3)/2 + s^2

    A = (sqrt(3) / 4) * t^2 + s^2

    3t + 4s = 20

    3t = 20 - 4s

    t = (4/3) * (5 - s)

    A = (sqrt(3)/4) * (4/3)^2 * (5 - s)^2 + s^2

    A = (sqrt(3)/4) * (16/9) * (25 - 10s + s^2) + s^2

    A = (16 * sqrt(3) / 36) * (25 - 10s + s^2) + s^2

    A = (4 * sqrt(3) / 9) * (25 - 10s + s^2) + s^2

    dA/ds = (4 * sqrt(3) / 9) * (0 - 10 + 2s) + 2s

    dA/ds = 0

    0 = (4 * sqrt(3) / 9) * (2s - 10) + 2s

    0 = 4 * sqrt(3) * (2s - 10) + 18s

    0 = 2 * sqrt(3) * (2s - 10) + 9s

    0 = 4 * sqrt(3) * s - 20 * sqrt(3) + 9s

    20 * sqrt(3) = s * (4 * sqrt(3) + 9)

    s = 20 * sqrt(3) / (9 + 4 * sqrt(3))

    s = 20 * sqrt(3) * (9 - 4 * sqrt(3)) / (81 - 16 * 3)

    s = 20 * sqrt(3) * (9 - 4 * sqrt(3)) / (81 - 48)

    s = 20 * sqrt(3) * (9 - 4 * sqrt(3)) / 33

    s = (20 * sqrt(3) / 33) * (9 - 4 * sqrt(3))

    Plug that back into 4s + 3t = 20 to get the value for t

    In case you hadn't figured it out, t is the side of the triangle and s is the side of the square

  • david
    Lv 7
    4 years ago

    s = length of 1 side of the square

    t = length of 1 side of the eq. triangle

    4s + 3t = 20

    t = (20 - 4s)/3

    Area of square = s^2

    area of eq. triangle = (1/2)[sq rt (3) t/2][t] = t^2 sq rt(3) / 4

    total A = s^2 + t^2 sq rt(3) / 4

    A(s) = s^2 + [(20 - 4s)/3]^2 sq rt(3) / 4

    A(s) = s^2 + [400 - 140s + 16s^2]sq rt(3) / 4

    A(s) = s^2 + 100 sq rt(3) - 35 s sq rt(3) + 4s^2 sq rt(3)

    A '(s) = 2s - 35 sq rt(3) + 8s sq rt(3) = 0

    [2 + 8 sqrt(3)] s = 35 sq rt(3)

    s = [ 35 sq rt(3) ] / [2 + 8 sqrt(3)] <<< side of the square

    t = (20 - 4s)/3 <<< sub value for s and find length of side of triangle

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