Posted by **nesri** on Tuesday, October 5, 2010 at 8:07am.

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

- calculus -
**Reiny**, Tuesday, October 5, 2010 at 9:28am
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)

- Correction - calculus -
**Reiny**, Tuesday, October 5, 2010 at 9:45am
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, ...)

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