Question

The area of a trapezium is defined by function f and given by f(x) = `(10 + "x") sqrt(100 - "x"^2)`, then the area when it is maximised is:

पर्याय

• 75 cm²

• `7sqrt3 "cm"^2`

• `75sqrt3 "cm"^2`

• 5 cm²

Answer 1

`75sqrt3 "cm"^2`

Explanation:

f'(x) = `(-2"x"^2 - 10"x" + 100)/(sqrt(100 - "x"^2))`

f'(x) = 0 ⇒ x = −10 or 5, But x > 0 ⇒ x = 5

f''(x) = `(2"x"^3 - 300"x" - 1000)/(100 - "x")^(3/2)`

⇒ f''(5) = `(-30)/sqrt75 < 0`

⇒ Maximum area of trapezium is `75sqrt3 "cm"^2` when x = 5