15. Among all triangles with the same base and the same angle at the vertex, find the triangle with the maximum perimeter.
15. Construct a segment on the segment $b$ that accommodates the angle $\beta$.

$$
\begin{gathered}
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} ; 2 p=a+b+c \\
2 p=\frac{b \sin A}{\sin B}+b+\frac{b \sin (A+B)}{\sin B}=b+\frac{b}{\sin \frac{B}{2}} \sin \left(A+\frac{B}{2}\right)
\end{gathered}
$$

Since $b>0, \sin \frac{B}{2}>0$, then $2 p$ will take the maximum value when $A+\frac{B}{2}=\frac{\pi}{2}$. In this case $\angle A=\angle C, \triangle A B C$ is isosceles.