Question

Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm

Answer 1

Given radius of the semi-circle = ‘r’ cm

Let the length of the rectangle be ‘x’ cm

Let the breadth of the rectangle be ‘y’ cm

From the figure,

`(x/2)^2 + y^2` = r²   ......[Pythagoras Theorem]

y² = `"r"^2 - x^2/4`

y = `1/2 sqrt(4"r"^2 - x^2)`

Area of the rectangle 'A' = xy

A = `x/2 sqrt(4"r"^2 - x^2)`

`"dA"/("d"x) = 1/2[(x(- 2x))/(2sqrt(4"r"^2 - x^2)) + sqrt(4"r"^2 - x^2)(1)]`

= `1/2[(4"r"^2 - 2x^2)/sqrt(4"r"^2 - x^2)]`

For maximum or minimum,

`"dA"/("d"x)` = 0

⇒ `1/2 [(4"r"^2  2x^2)/sqrt(4x^2 - x^2)]` = 0

4r² – 2x² = 0

x² = 2r²

x = `+-  sqrt(2)"r"`

x = `- sqrt(2) "r"` is not possible

∴ x = `sqrt(2) "r"`

Now, `("d"^2"A")/("d"x^2) = (sqrt(4"r"^2 - x^2) (- 2x) - (2"r"^2 - x^2)(- (2x)/(2sqrt(4"r"^2 - x^2))))/(4"r"^2 - x^2)`

= `(x(x^2 - 6"r"^2))/((4"r"^2 - x^2)sqrt(4"r"^2 - x^2))`

At x = `sqrt(2) "r", ("d"^2"A")/("d"x^2) < 0`

∴ Area of the rectangle is maximum

When x = `sqrt(2) "r"`

y = `1/2 sqrt(4"r"^2 - x^2)`

= `1/2 sqrt(4"r"^2 - 2"r"^2)`

= `1/2 sqrt(2"r"^2)`

= `(sqrt(2)"r")/2`

= `"r"/sqrt(2)`

∴ Length of the rectangle is `sqrt(2)` r cm

Breadth of the rectangle is `"r"/sqrt(2)` cm