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November 24, 2014

November 24, 2014

Posted by **Anonymous** on Thursday, January 24, 2008 at 9:54pm.

- Calculus -
**Reiny**, Thursday, January 24, 2008 at 10:32pmLet A(a,12-a^2) be the point of the triangle which lies on the parabola.

Since it must be an isosceles triangle, c = 2a

The area of the triangel = 1/2(c)(12-a^2)

=1/2(2a)(12-a^2)

=12a - a^3

Then d(Area)/da = 12 - 3a^2

= 0 for a max area

3a^2 = 12

a = +- 2, c = 2a

so c = 4

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