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Question
Show that the rectangle of the maximum perimeter which can be inscribed in the circle of radius 10 cm is a square of side `10sqrt2` cm.
Solution
`AB = 2 xx r cos theta`, `AD = 2r sin theta`
here r = 10
`AB = 20 cos theta, AD = 20 sin theta`
`A = 2rcos theta.2r sin theta` {Here r = 10}
`A= 4r^2 cos theta.sin theta`
`= 2r^2. 2sin theta cos theta`
`= 2 xx 100 sin 2 theta`
`(dA)/(d theta) = 200 cos 2 theta . 2`
`(dA)/(d theta) = 0`
`=> cos 2 theta = 0`
`2 theta = pi/2`
`theta = pi/4`
`(dA^2)/(d theta^2) = 400 (-2 sin 2 theta)`
`(dA^2)/(d theta^2) = -800 sin pi/2 = - 800< 0`
So, Area is max, when `AB = 20 cos pi/4 = 10sqrt2`
`AD = 20 cos pi/4 = 10sqrt2`
`:. AB = AD = 10sqrt2`
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