Question

an isosceles triangle has a vertex at the origin. Determine the area of the largest such triangle that is bound by the function x squared + 6y = 48 and explain why this is infact the maximum area.

Answer 1

because of the symmetry of the parabola, the base of the triangle must be parallel to the x-axis if the two sides are to be equal to form an isosceles triangle.

let the point of contact in the first quadrant be (x,y)

so the base of the triangle is 2x and its height is y
Area = 2xy
= 2x(8 - x^2/6)
= 16x - (1/3)x^3
d(Area)/dx = 16 - x^2= 0 for a max of area
x^2 = 16
x = ±4 ( and y = 16/3 )

so the largest area is
16(4) - 64/3 = 128/3 or 42 2/3

( I am assuming that the triangle lies above the x-axis. If you want the triangle to be located below the x-axis, of course the area would be without limits.
Make a sketch to understand this situation.