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Math Help - Optimization

  1. #1
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    Optimization

    A wire 60 cm long is to be cut into 2 pieces. One of the pieces will be bent into the shape of a square and the other into an equilateral triangle. The wire is to be cut in order that the sum of the areas of the square and the triangle is to be a maximum. An equation that can be used to model the total area of the shapes is:

    a(x) = x^2 / 16 + sqrt(3(60 - x)^2 / 18

    Determine the boundaries and the corresponding areas.

    x equals the perimeter of the square
    x - 60 equals the perimeter of the rectangle
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  2. #2
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    Quote Originally Posted by steezin View Post
    A wire 60 cm long is to be cut into 2 pieces. One of the pieces will be bent into the shape of a square and the other into an equilateral triangle. The wire is to be cut in order that the sum of the areas of the square and the triangle is to be a maximum. An equation that can be used to model the total area of the shapes is:

    a(x) = x^2 / 16 + sqrt(3(60 - x)^2 / 18

    Determine the boundaries and the corresponding areas.

    x equals the perimeter of the square
    x - 60 equals the perimeter of the rectangle
    Hi steezin,

    i guess you meant triangle.
    Also, this is a U-shaped quadratic in x, hence we can solve for minimum area, not maximum, using the derivative.
    Differentiating this wrt x and equate the result to zero, gives x corresponding to minimum area.
    You will see that it's U-shaped when you multiply out the factor on the RHS of the area equation.
    Hence, the maximum area corresponds to x=0 or x=60

    Sum of the areas = \frac{x^2}{16}+\frac{\sqrt{3}(60-x)^2}{18}

    For maximum area, compare the area when x=0 to the area when x=60.
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