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January 30, 2015

January 30, 2015

Posted by **Sani** on Wednesday, July 4, 2012 at 3:38am.

- maths -
**Reiny**, Wednesday, July 4, 2012 at 12:53pmplace your sketch on the x-y grid with the centre at (0,0)

let the radius of the semicircle be r, where r is a constant

so the base of the rectangle is 2x, let its height be y

then x^2 + y^2 = r^2

y =(r^2 - x^2)^(1/2)

area = 2xy

= 2x(r^2-x^2)^(1/2)

d(area)/dx = (2x)(1/2)(r^2 - x^2)^(-1/2) (-2x) + 2(r^2-x^2)^(1/2)

= 0 for a max/min of area

2(r^2 - x^2)^(1/2) = 2x^2/(r^2-x^2)^(1/2)

x^2 = r^2 - x^2

2x^2 = r^2

x^2 = r^2/2

x = r/√2 = (√2/2)r

then largest area

= 2x√(r^2 - x^2)

= (4√2/2)r (r^2 - r^2/2^(1/2)

= 2√2r(r/√2) = 2r^2

or, in a real simple way

suppose we look at the whole circle, the largest "rectangle we can fit inside the circle is a square, where x = y

then the sides of the square are 2x and 2y,

and the area is 4xy, but x=y and x^2 + x^2 = r^2 ---> x^2 = r^2/2

so the largest area = 4x^2

= 4(r^2/2) = 2r^2

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