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April 1, 2015

April 1, 2015

Posted by **Anonymous** on Tuesday, December 7, 2010 at 9:51pm.

- calculas -
**Reiny**, Tuesday, December 7, 2010 at 10:06pmDid you make a sketch?

draw a line from the centre to a vertex of the rectangle lying on the semicircle.

Let the length of the rectangle be 2x and its height y

I see a right-angled triangle with sides x and y and hypotenuse 2

x^2 + y^2 = 4

let the area be A

A = 2xy

A^2 = 4x^2y^2 = 4x^2(4-x^2)

= 16x^2 - 4x^4

2A dA/dx = 32x - 16x^3 = 0 for max of A

16x(2 - x^2) = 0

x^2=2

x=√2

sub back into equation above

y = √2

so length = 2√2 and width = √2

Perimeter = 6√2

- calculas -
**Anonymous**, Tuesday, December 7, 2010 at 10:10pmthank you so muchh :)

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