Posted by **Marissa** on Wednesday, June 1, 2011 at 1:10am.

A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y= 12 - x^2. What are the dimensions of such a rectangle with the greatest possible area?

- Calculus -
**Mgraph**, Wednesday, June 1, 2011 at 2:24am
The vertices of the rectangle are points:

(-x,0),(x,0),(x,12-x^2),(-x,12-x^2).

The area A(x)=2x(12-x^2)

A(x)=24x-2x^3

A'(x)=24-6x^2=0 => x=+-2

max A(x)=A(2)=48-16=32

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