Find the dimensions of the rectangle of largest area that has its base on the $\mathrm{x}$-axis and its other two vertices above the $x$-axis and lying on the parabola $y=16-x^{2}$. List the dimensions in non-decreasin order.

Step 1 ：The problem is asking for the dimensions of the rectangle with the largest area that fits under the parabola \(y=16-x^{2}\) and has its base on the x-axis. The rectangle's other two vertices are on the parabola.

Step 2 ：We can express the area of the rectangle as a function of x, then find the maximum of this function. The width of the rectangle is \(2x\) (since it extends from -x to x on the x-axis), and the height is the y-value of the parabola at x, which is \(16-x^{2}\). So the area A of the rectangle as a function of x is \(A(x) = 2x(16-x^{2})\).

Step 3 ：We can find the maximum of this function by taking its derivative, setting it equal to zero, and solving for x. The x-values that satisfy this equation are the x-coordinates of the points where the function has a local maximum or minimum. We can then substitute these x-values back into the function to find the corresponding y-values, which give us the dimensions of the rectangle.

Step 4 ：The critical points of the area function are \(x = -\frac{4\sqrt{3}}{3}\) and \(x = \frac{4\sqrt{3}}{3}\). These are the x-coordinates of the points where the area of the rectangle is maximized.

Step 5 ：The corresponding dimensions of the rectangle are \((\frac{8\sqrt{3}}{3}, \frac{32}{3})\) and \((\frac{8\sqrt{3}}{3}, \frac{32}{3})\). Since the question asks for the dimensions in non-decreasing order, we can see that the dimensions are already in non-decreasing order.

Step 6 ：Final Answer: The dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola \(y=16-x^{2}\) are \(\boxed{\left(\frac{8\sqrt{3}}{3}, \frac{32}{3}\right)}\).