Calculus

An isosceles triangle has a vertex at the origin. Determine the area of the largest such triangle that is bound by the function x^2 + 6y = 48.

So, y=8 when x=0 making a vertical side equal to 8. I also determined that there is a zero at =/- the square root of 48, or approximately 6.93. I know that the base of a triangle=bh/2. I also know that two of the sides must be equal. But, I'm not sure where to go from here. Please help !!

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  1. Did you make a sketch ?

    The parabola is y = -(1/6)x^2 + 8 or -x^2/6 + 8
    Let P(x,y) and Q(-x,y) be the base of the triangle, both P and Q on the parabola in the first and second quadrants.

    The the area of the triangle is
    A = (1/2)(2x)(y) = xy
    = x(-x^2/6 + 8)
    = -x^3/6 + 8x

    d(A)/dx = (-1/2)x^2 + 8 = 0 for a max of A
    x^2 = 16
    x = ± 4

    so Area = -x^3/6 + 8x
    = -64/6 + 32
    = 64/3 units^2

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