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

An isosceles triangle, whose base is the interval from (0,0) to (c,0), has its vertex on the graph of f(x)=12-x^2. For what value of c does the triangle have maximum area?

- Calculus -
**Reiny**, Thursday, January 24, 2008 at 10:32pm
Let 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

## Answer this Question

## Related Questions

- calculus - 5. Let for and f(x)=12X^2 for x>=0 and f(x)>=0 a. The line ...
- calculus - An isosceles triangle is drawn with its vertex at the origin and its ...
- Math - An isosceles triangle has its vertex at the origin and the ends of its ...
- Calculus - A right triangle has one leg on the x-axis. The vertex at the right ...
- corollary to isosceles triangle theorem - how to prove 1)the measure of each ...
- math - 9. An isosceles triangle is drawn with its vertex at the origin and its ...
- Calculus - An isosceles triangle has a vertex at the origin. Determine the area ...
- math - an isosceles triangle has a vertex at the origin. Determine the area of ...
- Precalculus - The sides of an isosceles triangle are 10 cm, 10 cm, and 12 cm. A ...
- Calculus - AN ISOSCELES TRIANGLE IS TO HAVE A PERIMETER OF 64CM. DETERMINE THE ...