What is the maximum area of the triangle?

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,

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

To maximize, differentiate w.r.t. \(x\) and set to zero:

\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} \;-\;\frac{6x}{2\sqrt{9-6x}} = 0 \;\Longrightarrow\; x = 1 \)

At \(x = 1\), we get \(BC = \sqrt{9 - 6} = \sqrt{3}\), so

\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)

∴ The maximum area is \(\frac{\sqrt{3}}{2}\) square unit.

Hence, the correct answer is Option  1.