calculus
posted by nesri .
An isosceles triangle is drawn with its vertex at the origin and its base parallel to the xaxis. The vertices of the base are on the curve 5y=25x^2 Find the area of the largest such triangle.

your parabola is y = 5  (1/5)x^2
let the points of contact of the base of the triangle be (x,y) and (x,y)
Then the area is
Area = xy/2
= (x/2)(5(1/5)x^2)
= 5x/2  (1/10)x^3
d(Area)/dx = 5/2  3x^2/10
= 0 for a max of Area
3x^2/10 = 5/2
x^2 = 25/3
x = ± 5/√3 or appr. 2.88675
sub back in area expression
I had appr. 4.81125
( I tried x = 3 and x = 2.8 and they both gave me slightly smaller areas) 
The base of the triangle should have been 2x, which makes the
Area = xy
= 5x  (1/5)x^3
Can you figure out how that changes the answers?
(hint: one stays the same, ...)