Pizza time.

Let the radius of the sector be $r$ and its central angle be $\theta$ . Then the area of the sector $A = \dfrac \theta 2 r^2$ and its perimeter $p = 2r+\theta r = 20$ . Then we have:

$\begin{aligned} 2r + \theta r & = 20 & \small \color{#3D99F6} \text{Rearranging} \\ \theta r & = 20 - 2r & \small \color{#3D99F6} \text{Multiply both sides by }\frac r2 \\ \frac \theta 2 r^2 & = 10r - r^2 \\ A & = 25 - (r^2 - 10r + 2) \\ & = 25 - (r-5)^2 & \small \color{#3D99F6} \text{Since }(r-5)^2 \ge 0 \\ \implies A & \le \boxed{25} \end{aligned}$

The perimeter of the sector is $2r+rθ=20$ , so we get $r=\frac{20}{θ+2}$ . Substituting this into the sector area formula, we get $A(θ)=\frac{200θ}{(θ+2)^2}$ . Differentiating and setting $A'(θ)=0$ , we yield $θ=2$ .

Substituting this back into $A(θ)$ , we get the maximum area as $\frac{400}{16}=25m^2$

To prove that this is a maximum, we can find the second derivative, $A''(θ)=\frac{400(θ-4)}{(θ+2)^4}$ At $θ=2$ , the value of this is $A''(2)=-3.125$ . Since this is negative, a maximum occurs at $θ=2$ .

Let the radius of the sector be $r$ and the central angle of the sector be $\theta$ . Then the perimeter of the sector is $2r + r\theta = 20 \Longrightarrow \theta = \dfrac{20 - 2r}{r}$ .

The area of the sector is then $\dfrac{1}{2}r^{2} \theta = \dfrac{1}{2} r^{2} \times \dfrac{(20 - 2r)}{r} = r(10 - r)$ . Now by the AM-GM inequality

$2\sqrt{r(10 - r)} \le r + (10 - r) = 10 \Longrightarrow \sqrt{r(10 - r)} \le 5 \Longrightarrow r(10 - r) \le 25$ , and thus the maximum possible area is $\boxed{25}$ ,

which occurs when $r = 10 - r \Longrightarrow r = 5 \Longrightarrow \theta = 2$ .

Let $r$ be the radius of the sector and $s$ be its arc length. Then its perimeter $P = 2r - s = 20$ and its area $A = \frac{1}{2}rs$ .

Combining these equations gives $A = \frac{1}{2}r(20 - 2r) = 10r - r^2 = 25 - (r - 5)^2$ , which means $A \leq 25$ , so the maximum area is $A = \boxed{25}$ .