Let ABC be a triangle right-angled at B and AB + AC = 3 units. If the area of the triangle is maximum, then ∠A is equal to:

Let AB = c, AC = b, with c + b = 3 and BC = √(b²-c²). Area = (1/2)c·√(b²-c²). With c = 3-b: Area² = (1/4)(3-b)²(b²-(3-b)²) = (1/4)(3-b)²(b-3+b)(b+3-b)·... = (1/4)(3-b)²(2b-3)(3). Maximize: take derivative w.r.t. b. Setting d/db[(3-b)²(2b-3)] = 0: -2(3-b)(2b-3) + 2(3-b)² = 0, (3-b)[-2(2b-3) + 2(3-b)] = 0, -(4b-6) + (6-2b) = 0, -4b+6+6-2b = 0, -6b+12 = 0, b = 2. So AC = 2, AB = 1, cosA = AB/AC = 1/2, A = π/3.