What is ∠A equal to if the area of the triangle is maximum?

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x\).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,\sqrt{9 - 6x} \)

To maximize, set up

\(A^2 = \tfrac14\,x^2\,(9 - 6x)\)

\(\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 \)

Hence \(x = 1\) (discarding x=0, so

\(BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2\)

Therefore,

\(\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}\)

∴ \(\angle A = \frac{\pi}{3}\).

Hence, the correct answer is Option 3.