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Question
Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.
Solution
Let ABCD be the rectangle inscribed in a semicircle of radius 1 unit such that the vertices A and B lie on the diameter.
Let AB = DC = x and BC = AD = y.
Let O be the centre of the semicircle.
Join OC and OD. Then OC = OD = radius = 1.
Also, AD = BC and m∠A = m∠B = 90°.
∴ OA= OB
∴ OB = `(1)/(2) "AB" = x/(2)`
In right angled triangle OBC,
OB2 + BC2 = OC2
∴ `(x/2)^2 + y^2` = 12
∴ y2 = `1 - x^2/(4) = (1)/(4)(4 - x^2)`
∴ y = `(1)/(2)sqrt(4 - x^2)` ...[∵ y > 0]
Also of the triangle
= xy
= `x.(1)/(2)sqrt(4 - x^2)`
Let f(x) = `(1)/(2) xx sqrt(4 - x^2)`
= `(1)/(2)sqrt(4x2 - x^4)`
∴ f'(x) = `(1)/(2)d/dx(sqrt(4x^2 - x^4))`
= `(1)/(2) xx (1)/(2sqrt(4x^2 - x^4)) xx.d/dx(4x^2 - x^4)`
= `(1)/(4sqrt(4x^2 - x^4)) xx (4 xx 2x - 4x^3)`
= `(4x(2 - x2))/(4xsqrt(4 - x^2)`
= `(2 - x^2)/sqrt(4 - x^2)` ...[∵ x ≠ 0]
and
f"(x) = `d/dx((2 - x^2)/sqrt(4 - x^2))`
= `d/dx[((4 - x^2) - 2)/sqrt(4 - x^2)]`
= `d/dx[sqrt(4 - x^2) - (2)/sqrt(4 - x^2)]`
= `d/dx(sqrt(4 - x^2)) - 2d/dx(4 - x^2)^(-1/2)`
= `(1)/(2sqrt(4 - x^2)).d/dx(4- x^2) - 2(-1/2)(4 - x^2)^(-3/2).d/dx(4 - x^2)`
= `(1)/(2sqrt(4 -+ x^2)) xx (0 - 2x) + (1)/(4 - x^2)^(3/2) xx (0 - 2x)`
= `-x/sqrt(4 - x^2) - (2x)/(4 - x^2)^(3/2)`
= `(-x(4 - x^2) - 2x)/(4 - x^2)^(3/2)`
= `(-4x + x^3 - 2x)/(4 - x^2)^(3/2)`
= `(x^3 - 6x)/(4 - x^2)^(3/2)`
For maximum value of f(x),f'(x) = 0
∴ `(2 - x^2)/sqrt(4 - x^2)` = 0
∴ 2 – x2 = 0
∴ x2 = 2
∴ x = `sqrt(2)` ...[∵ x > 0]
Now, f"`(sqrt(2)) = ((sqrt(2))^3 - 6sqrt(2))/[4 - (sqrt(2))^2]^(3/2)`
= `(-4sqrt(2))/(2sqrt(2))`
= – 2 < 0
∴ by the second derivative test, f is maximum when x = `sqrt(2)`
When `x = sqrt(2), y = (1)/(2)sqrt(4 - x^2)`
= `(1)/(2)sqrt(4 - 2)`
= `(1)/(2) xx sqrt(2)`
= `(1)/sqrt(2)`
∴ `x = sqrt(2) and y = (1)/sqrt(2)`
Hence, the area of the rectangle is maximum (i.e. rectangle has the largest size) when its length is `sqrt(2)` units and breadth is `(1)/sqrt(2)`unit.
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