Let the length and width of the rectangle be x and y, respectively.

The perimeter of the rectangle is given by \(p = 2x + 2y\).

Solving for y, we get \(y = \frac{p}{2} - x\).

The area of the rectangle is given by \(A = xy = x\left(\frac{p}{2} - x\right) = \frac{px}{2} - x^2\).

To find the dimensions of the rectangle of the greatest area, we need to maximize the area with respect to x.

Taking the derivative of the area with respect to x, we get \(\frac{dA}{dx} = \frac{p}{2} - 2x\).

Setting this equal to zero and solving for \(x\), we find that \(x = \frac{p}{4}\).

Substituting this value into the expression for \(y\), we get \(y = \frac{p}{2} - x = \frac{p}{4}\).

So, the dimensions of the rectangle of the greatest area are both \(\frac{p}{4}\), which means that the rectangle is a square.