Question

Solve the following : Show that of all rectangles inscribed in a given circle, the square has the maximum area.

Answer 1

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.