Find the dimensions of the largest rectangle that can be inscribed in the semicircle y = `sqrt(4-x^2)` , such that the areas of the rectangle is a maximum.
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Let ABCD is rectangle inscribed in circle in such a way that side CD is along diameter of the circle.Let O be the centre of the circle and bisect side CD. Join O to B and further let `angle OBC=theta`
In triangle OBC, OB=radius of the semicircle=2
Area A of the rectangle ABCD= BC x CD
A = `2cos(theta)xx(2 xx2sin(theta))`
for max / min
`` Thus `theta=pi/4` ,will give maxmum area.
Thus dimension of rectangle
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