What is \angle A equal to if the area of the triangle is <strong>MAXIMUM</strong>?

Consider the following for the two (02) items that follow : Let ABC be a triangle right-angled at B and AB+AC=3 units.

  1. A. \pi/6
  2. B. \pi/4
  3. C. \pi/3
  4. D. 5\pi/12

Correct Answer: C. \pi/3

Explanation

Let AB = c, AC = b, BC = a. Given c + b = 3 \implies b = 3 - c. By Pythagoras theorem, a^2 + c^2 = b^2 = (3-c)^2 = 9 - 6c + c^2 \implies a^2 = 9 - 6c. The area is \Delta = \frac{1}{2}ac. To maximize \Delta, we maximize \Delta^2 = \frac{1}{4}c^2(9-6c) = \frac{1}{4}(9c^2 - 6c^3). Differentiating wrt c and equating to 0: 18c - 18c^2 = 0 \implies c = 1. Then b = 3 - 1 = 2. Therefore, \cos A = \frac{c}{b} = \frac{1}{2} \implies \angle A = \frac{\pi}{3}.

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