Posted by **Patrick** on Thursday, November 1, 2012 at 1:23am.

Find the dimensions of the rectangle with the largest area that is inscribed inside the parabola y = 16- x^2 and the x-axis

- calculus -
**Steve**, Thursday, November 1, 2012 at 10:09am
since we want a rectangle, the top side to be parallel to the bottom side. So, the rectangle is centered over (0,0).

Let the base of the rectangle extend from -x to x

The area is thus

a = 2xy = 2x(16-x^2)

da/dx = 32 - 6x^2

da/dx=0 when x = 4/√3

the rectangle is thus 8/√3 by 32/3

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