If $a$,$b$,$b$ are the sides of the isosceles triangle, the height of the triangle $h$ would be,$$h=\sqrt{b^2-\left(\frac a 2\right)^2}$$
The perimeter $C$ and the area $A$ are given as,
$$C=a+2b\qquad (1)$$
$$A=\frac{a\times h}{2}=\frac{a\times \sqrt{b^2-\left(\frac a 2\right)^2}}{2}=\frac{a\sqrt{4b^2-a^2}}{4}$$
From (1),
$$\begin{align*}
A&=\frac{a\sqrt{4\left(\frac{C-a}{2}\right)^2-a^2}}{4}\\
&=\frac{a\sqrt{(C-a)^2-a^2}}{4}\\
&=\frac{a\sqrt{(C-a+a)(C-a-a)}}{4}\\
&=\frac{a\sqrt{C(C-2a)}}{4}\\
&=\frac{\sqrt{C(Ca^2-2a^3)}}{4}
\end{align*}$$
$$A^2=\frac{C(Ca^2-2a^3)}{16}$$
by differentiating with respect to a,
$$\begin{align*}
2A\frac{dA}{da}&=\frac{C(2aC-6a^2)}{16}\\
\frac{dA}{da}&=\frac{C(2aC-6a^2)}{32A}
\end{align*}$$
To find max/min $\frac{dA}{da}=0$
$$\frac{2aC-6a^2}{32A}=0$$
since $A>0$,
$$\begin{align*}
2aC-6a^2&=0\\
a(C-3a)&=0\\
a&=\frac{C}{3} \quad (\text{since $a\not=0$})
\end{align*}$$
by taking, second derivative,
$$2\left(\frac{dA}{da}\right)^2+2A\frac{d^2A}{da^2}=\frac{2C-12a}{16}$$
when $\frac{dA}{da}=0$, $a=\frac{C}{3}$
$$\begin{align*}
2(0)+2A\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}&=\frac{C-2C}{8}\\
\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}&= \frac{-C}{16A}\\
\end{align*}$$
with $\text{$C>0$ and $ A>0$}$
$$\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}<0$$
Therefore the point $a=\frac{C}{3}$ is a maximum.
Therefore the area becomes maximum when $a=\frac{C}{3}$
from (1),
$$3a=a+2b\implies a=b$$
therefore the area of the isosceles triangle becomes maximum when when the triangle is an equilateral triangle with a given constant perimeter
