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Question
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`
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Solution
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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