Among triangles with a given base and perimeter, the isosceles one is the largest

Heron. The semi-perimeter $\sigma=\frac{b+s}2$ is constant. The sides are $b$, $x$, $s-x$, hence
$$A^2=\sigma(\sigma-b)\cdot(\sigma-x)\bigl(\sigma-(s-x)\bigr).$$
Since $A>0$, maximising $A$ is equivalent to maximising $A^2$.

The variable part. With $u=x-\frac s2$, we have $\sigma-x=\frac b2-u$ and $\sigma-(s-x)=\frac b2+u$, so
$$(\sigma-x)\bigl(\sigma-s+x\bigr)=\frac{b^2}4-u^2.$$
The triangle inequalities hold exactly when $|u|<\frac b2$, and on this range the expression above is maximal at $u=0$, i.e. $x=\frac s2$: the isosceles triangle.

Value. $\sigma(\sigma-b)=\dfrac{(s+b)(s-b)}4=\dfrac{s^2-b^2}4$, so
$$A_{\max}^2=\frac{s^2-b^2}4\cdot\frac{b^2}4,\qquad A_{\max}=\frac b4\sqrt{s^2-b^2}.$$

Remark. Geometrically: the apex moves on an ellipse with foci at the endpoints of the base (the sum of the distances $s$ is constant), and the highest point of this ellipse is above the middle of the base.