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

Thus BC=2`cos(theta)`

`OC=2sin(theta)`

Area A of the rectangle ABCD= BC x CD

A = `2cos(theta)xx(2 xx2sin(theta))`

`A=4sin(2theta)`

`A'=8cos(2theta)`

for max / min

A'=0

`2theta=pi/2`

`theta=pi/4 `

`A''=-16sin(2theta)`

`A''}_{theta=pi/4}=-16 <0`

`` Thus `theta=pi/4` ,will give maxmum area.

Thus dimension of rectangle

BC=`2 cos(pi/4)=2xx1/(sqrt(2))=sqrt(2)`

`CD=2xx 2sin(pi/4)=4xx(1/sqrt(2))`

`=2sqrt(2)`

Area=4 sq.unit

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