Prove the identity (sinx + cosx)(1-sinxcosx) = sin³x + cos³x.

The perimeter of the sector POQ is the sum of the two radii and the arc length.
Perimeter $P = OP + OQ + \text{arc PQ}$.
Given $P = 50 \text{ cm}$, radius $OP = OQ = r \text{ cm}$, and angle $POQ = \theta$ radians.
The arc length $PQ = r\theta$.
So, $50 = r + r + r\theta$

$$50 = 2r + r\theta$$

Rearrange to express $\theta$:

$$r\theta = 50 - 2r$$ $$\theta = \frac{50 - 2r}{r} = \frac{50}{r} - 2$$

The area of a sector is given by $A = \frac{1}{2}r^2\theta$.
Substitute $\theta = \frac{50 - 2r}{r}$:

$$A = \frac{1}{2}r^2 \left(\frac{50 - 2r}{r}\right)$$ $$A = \frac{1}{2}r(50 - 2r)$$ $$A = 25r - r^2$$

Thus, the area of the sector is $\boxed{A = 25r - r^2}$.

Given $A = 25r - r^2$.

$$\frac{dA}{dr} = \frac{d}{dr}(25r - r^2) = 25 - 2r$$

Set $\frac{dA}{dr} = 0$ to find the stationary value of $r$:

$$25 - 2r = 0$$ $$2r = 25$$ $$r = \frac{25}{2} = 12.5$$

Substitute $r = 12.5$ into the area formula $A = 25r - r^2$:

$$A = 25(12.5) - (12.5)^2$$ $$A = 312.5 - 156.25$$ $$A = 156.25$$

$$\frac{d^2A}{dr^2} = \frac{d}{dr}(25 - 2r) = -2$$

Since $\frac{d^2A}{dr^2} = -2 < 0$, the stationary point corresponds to a maximum value.
The stationary value of $A$ is $\boxed{156.25 \text{ cm}^2}$, and its nature is a maximum.