Calculus
posted by Jeff .
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 !!

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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