Posted by **Jeff** on Saturday, March 20, 2010 at 8:08pm.

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 !!

- Calculus -
**Reiny**, Saturday, March 20, 2010 at 8:35pm
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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