Calculus

An isosceles triangle, whose base is the interval from (0,0) to (c,0), has its vertex on the graph of f(x)=12-x^2. For what value of c does the triangle have maximum area?

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  1. Let A(a,12-a�^2) be the point of the triangle which lies on the parabola.
    Since it must be an isosceles triangle, c = 2a

    The area of the triangel = 1/2(c)(12-a^2)
    =1/2(2a)(12-a^2)
    =12a - a^3

    Then d(Area)/da = 12 - 3a^2
    = 0 for a max area

    3a^2 = 12
    a = +- 2, c = 2a
    so c = 4

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    posted by Reiny

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