$A = \sqrt{s(s - a)(s - b)(s - c)}$
Where
$P = \text{given}$
$s = \frac{1}{2}P$
$a + b + c = P$
$c = P - a - b$ ← equation (1)
Thus,
$A = \sqrt{\frac{1}{2}P(\frac{1}{2}P - a)(\frac{1}{2}P - b)[\frac{1}{2}P - (P - a - b)]}$
$A = \sqrt{\frac{1}{2}P(\frac{1}{2}P - a)(\frac{1}{2}P - b)(a + b - \frac{1}{2}P)}$
$\dfrac{\partial A}{\partial a} = \dfrac{\frac{1}{2}P(\frac{1}{2}P - b)\left[ (\frac{1}{2}P - a)(1) + (a + b - \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P - a)(\frac{1}{2}P - b)(a + b - \frac{1}{2}P)}} = 0$
$\frac{1}{2}P - a = a + b - \frac{1}{2}P$
$b = P - 2a$ ← equation (2)
$\dfrac{\partial A}{\partial b} = \dfrac{\frac{1}{2}P(\frac{1}{2}P - a)\left[ (\frac{1}{2}P - b)(1) + (a + b - \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P - a)(\frac{1}{2}P - b)(a + b - \frac{1}{2}P)}} = 0$
$\frac{1}{2}P - b = a + b - \frac{1}{2}P$
$a = P - 2b$
Substitute b = P - 2a of equation (2)
$a = P - 2(P - 2a)$
$a = P - 2P + 4a$
$3a = P$
$a = \frac{1}{3}P$
From equation (2)
$b = P - 2(\frac{1}{3}P)$
$b = \frac{1}{3}P$
From equation (1)
$c = P - \frac{1}{3}P - \frac{1}{3}P$
$c = \frac{1}{3}P$
Hence, a = b = c = P/3. Thus, the triangle is equilateral. answer
The maximum area of the triangle is:
$A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{2}P - \frac{1}{3}P)(\frac{1}{2}P - \frac{1}{3}P)(\frac{1}{2}P - \frac{1}{3}P)}$
$A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{6}P)(\frac{1}{6}P)(\frac{1}{6}P)}$
$A_{max} = \sqrt{\frac{1}{432}P^4}$
$A_{max} = \frac{1}{12\sqrt{3}}P^2$ answer
