Task B-3.7. What is the maximum value of the area of a triangle with a perimeter of $20 \mathrm{~cm}$ and one side length of $8 \mathrm{~cm}$?
## First Solution.

Let the length of the side $b=8 \mathrm{~cm}$ and $a+b+c=20 \mathrm{~cm}$. We calculate the area of the triangle using Heron's formula

$$
P=\sqrt{s(s-a)(s-b)(s-c)}, \quad s=\frac{o}{2}=10
$$

From the perimeter, it follows that $c=20-8-a=12-a$. Then,

$$
\begin{aligned}
& P=\sqrt{10(10-a)(10-8)(a-2)} \\
& P=\sqrt{20\left(-a^{2}+12 a-20\right)}
\end{aligned}
$$

This expression will have its maximum value for the value of $a$ for which the quadratic function $f(a)=-a^{2}+12 a-20$ has its maximum value, which is at $a=6 \mathrm{~cm}$,

(2 points) Then $c=6 \mathrm{~cm}$, and the maximum area

(1 point) $P=8 \sqrt{5} \mathrm{~cm}^{2}$.

(2 points)