A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) Ho

Question

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire should be used for the square in order to maximize the total area? 29 m (b) How much wire should be used for the square in order to minimize the total area?

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Minh Khuê 4 years 2021-08-10T09:31:26+00:00 1 Answers 277 views 0

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  1. quangkhai
    -2
    2021-08-10T09:32:37+00:00
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    Answer:

    Explanation:

    Total length of the wire is 29 m.

    Let the length of one piece is d and of another piece is 29 – d.

    Let d is used to make a square.

    And 29 – d is used to make an equilateral triangle.

    (a)

    Area of square = d²

    Area of equilateral triangle = √3(29 – d)²/4

    Total area,

    A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

    Differentiate both sides with respect to d.

    \frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

    For maxima and minima, dA/dt = 0

    d = 8.76 m

    Differentiate again we get the

    \frac{d^{2}A}{dt^{2}}= + ve

    (a) So, the area is maximum when the side of square is 29 m

    (b) so, the area is minimum when the side of square is 8.76 m

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