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December 24, 2014

December 24, 2014

Posted by **jim** on Tuesday, May 19, 2009 at 8:06pm.

y=12−x2 .

What is the largest area that the rectangle can have

- calculus -
**Reiny**, Tuesday, May 19, 2009 at 8:16pmlet the points of contact be (x,y) and (-x,y)

so the base is 2x and the height is y

but y=12-x^2

so the

Area = 2xy

= 2x(12-x^2)

= 24x - 2x^3

d(Area)/dx = 24 - 6x^2

= 0 for a max Area

solve 2x^3 - 24 = 0

sub that back into the Area = 2x(12-x^2)

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