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Question
Solve the following : Show that of all rectangles inscribed in a given circle, the square has the maximum area.
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Solution
Let ABCD be a rectangle inscribed in a circle of radius r. Let AB = x and BC = y.
Then x2 + y2 = 4r2 … (1)
Area of rectangle = xy
= `xsqrt(4r^2 - x^2)` ...[By (1)]
Let f(x) = x2(4t2 – x2)
= 4r2x2 – x4
∴ f'(x) = `d/dx(4r^2x^2 - x^4)`
= 4r2 x 2x – 4x3
= 8r2x – 4x3
and
f"(x) = `d/dx(8r^2x - 4x^3)`
= 8r2 x 1 – 4 x 3x2
= 8r2 – 12x2
For maximum area, f'(x) = 0
∴ 8r2x – 4x3 = 0
∴ 4x3 = 8r2x
∴ x2 = 2r2 ...[∵ x ≠ 0]
∴ x = `sqrt(2)r` ...[∵ x > 0]
and
`f"(sqrt(2r)) = 8r^2 – 12(sqrt(2r))`
= – 16r2 < 0
∴ f(x) is maximum when x = `sqrt(2)r`
If x = `sqrt(2)r`, then from (1),
`(sqrt(2r))^2 + y^2` = 4r2
∴ y2 = 4r2 – 2r2 = 2r2
∴ y = `sqrt(2)r` ...[∵ y > 0]
∴ x = y
∴ rectangle is a square.
Hence, amongst all rectangles inscribed in a circle, the square has maximum area.
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