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Question
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
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Solution
Let ABCD be a rectangle inscribed in the given circle of radius r having centre at O.
Let one side of the rectangle be c then the other side
`= sqrt((2r)^2 - x^2)`
`= sqrt(4r^2 - x^2)`
(∴ ∠ADC = ∠ABC = 90°; an angle in the semi-circle).
Let A be the corresponding area of the rectangle.
`A = xsqrt(4r^2 - x^2), 0 < x < 2r`
⇒ `(dA)/dx = (x(-2x))/(2 sqrt(4r^2 - x^2)) + sqrt(4r^2 - x^2)`
`= (2(2r^2 - x^2))/sqrt(4r^2 - x^2)`
For maximum / minimum area
`(dA)/dx = (2r^2 - x^2)/(sqrt(4r^2 - x^2)) = 0`
⇒ `x = sqrt2r` ...(∵ 0 < x < 2r)
Now,
`(d^2A)/dx^2 = (sqrt(4r^2 - x^2) (-4x) - (4r^2 - 2x^2) (1xx(-2x))/(2sqrt(4r^2 - x^2)))/((4r^2 - x^2))`
`= ((4r^2 - x^2) (-4x) + (4r^2 - 2x^2) x)/((4r^2 - x^2)^(3//2))`
`= (-12r^2x + 2x^3)/(4r^2 - x^2)^(3//2)`
`((d^2A)/dx^2)_(x = sqrt(2r))`
`= (-12r^2 (sqrt (2r)) + 2 (sqrt (2r))^3)/((2r^2)^(3//2)`
`= (4sqrt(2r^3) - 12 sqrt (2r^3))/(2 sqrt(2r^3))`
= 2 - 6 < 0
∴ Area is maximum at x = `sqrt(2r)`
∴ length of rectangle is `sqrt(2r)`
width of rectangle is `sqrt (4r^2 - x^2) = sqrt (2r)`
Hence the rectangle is a square of side `sqrt (2r)` for maximum area.
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