Let ABC be a triangle right-angled at B and AB + AC = 3 units. What is ∠A equal to if the area of the triangle is maximum?

Let ∠A = θ. Then AB = AC cosθ (adjacent), and BC = AC sinθ. Given AB + AC = 3, so AC cosθ + AC = 3, hence AC = 3/(1 + cosθ). Area = (1/2)·AB·BC = (1/2)·AC cosθ·AC sinθ = (1/2)AC² sinθ cosθ = (1/4)AC² sin2θ. Substituting AC = 3/(1+cosθ): Area = (9/4)·sin2θ/(1+cosθ)². Differentiating and setting to zero leads to optimal θ = π/6, where the triangle area is maximized.