मराठी

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is π3

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प्रश्न

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is `pi/3`

बेरीज
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उत्तर


Let ΔABC be the right angled triangle in which ∠B = 90°

Let AC = x, BC = y

∴ AB = `sqrt(x^2 - y^2)`

∠ACB = θ

Let Z = x + y ....(Given)

Now area of ΔABC, A = `1/2 xx "AB" xx "BC"`

⇒ A = `1/2 y * sqrt(x^2 - y^2)`

⇒ A = `1/2 y * sqrt(("Z" - y)^2 - y^2)`

Squaring both sides, we get

⇒ A2 = `1/4 y^2 [("Z" - y)^2 - y^2]`

⇒ A2 = `1/4 y^2 ["Z"^2 + y^2 - 2"Z" y - y^2]`

⇒ P = `1/4 y^2 ["Z"^2 - 2"Z"y]`

⇒ P = `1/4 [y^2"Z"^2 - 2"Z"y^3]`  ....[A2 = P]

Differentiating both sides w.r.t. y we get

`"dP"/"dy" = 1/4 [2y"Z"^2 - 6"Z"y^2]`  .....(i)

For local maxima and local minima,

`"dP"/"dy"` = 0

∴ `1/4 (2y"Z"^2 - 6"Z"y^2)` = 0

⇒ `(2y"Z")/4 ("Z" - 3y)` = 0

⇒ yZ(Z – 3y) = 0

⇒ yZ ≠ 0   .....(∵ y ≠ 0 and Z ≠ 0) 

⇒ Z – 3y = 0

⇒ y = `"Z"/3`

⇒ y = `(x + y)/3`  .....(∵ Z = x + y)

⇒ 3y = x + y

⇒ 3y – y = x

⇒ 2y = x

⇒ `y/x = 1/2`

⇒ cos θ = `1/2`

∴  θ = `pi/3`

Differentiating eq. (i) w.r.t. y,

We have `("d"^2"P")/("dy"^2) = 1/4 [2"Z"^2 - 12"Z"y]`

`("d"^2"P")/("dy"^2)` at y = `"Z"/3 = 1/4 [2"Z"^2 - 12"Z" * "Z"/3]`

= `1/4 [2"Z"^2 - 4"Z"^2]`

= `(-"Z"^2)/2 < 0`

Hence, the area of the given triangle is maximum when the angle between its hypotenuse and a side is `pi/3`.

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पाठ 6: Application Of Derivatives - Exercise [पृष्ठ १३७]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 6 Application Of Derivatives
Exercise | Q 25 | पृष्ठ १३७

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