Sol. Let ABC be the triangle formed by folding the wire of length λ cm such that ∠ A = 60º, BC = a, CA = b and AB = c.

Then, a + b + c = λ

Now,

∠ A = 60º

⇒ cos A =

⇒ =

⇒ b 2 + c 2 –a 2 = bc

⇒ (b + c) 2 –a 2 = 3bc

⇒ ( λ –a) 2 –a 2 = 3bc

[  a + b + c = λ ; b + c = λ –a]

⇒ λ 2 –2a λ = 3bc ... (i)

Now,

bc = λ ( λ –2a) and b + c = λ – a

∴ (b – c) 2 = (b + c) 2 –4bc

⇒ (b – c) 2 = ( λ –a) 2 – λ ( λ –2a)

⇒ (b – c) 2 = (– λ 2 +2a λ + 3a 2 ) ... (ii)

 (b – c) 2 ≥ 0

∴ (– λ 2 + 2a λ + 3a 2 ) ≥ 0

⇒ λ 2 –2a λ –3a 2 ≤ 0

⇒ ( λ –3a) ( λ + a) ≤ 0

⇒ λ –3a ≤ 0 [  λ + a > 0]

⇒ λ ≤ 3a

⇒ a ≥ λ /3

⇒ a ∈ [ λ /3, λ ] ... (iii)

Let Δ be the area of Δ ABC. Then,

Δ = bc sin 60º

⇒ Δ = bc

⇒ Δ = bc

⇒ Δ = ( λ 2 –2a λ ) [using (i)]

⇒ = (–2 λ ) < 0

⇒ Δ is a decreasing function of a.

⇒ Δ is maximum for a =

[  a ∈ [ λ /3, λ ) from (iii)]

Putting a = in (ii), we get b = c

Substituting a = and b = c in a + b + c = λ , we get

a = b = c =

Hence, area is maximum when sides of the triangle are equal and each equal to λ /3.