Solution: First, we can see that the base of the rectangle is made up of two \(x\)'s. This is because \(x\) only measures from the origin to the value on the right. Secondly, the height of the rectangle must be determined by the formula \(y=6-x^2\). Thus, the area of the rectangle is given by \(A=2x(6-x^2)\) whereas the perimeter is given by \(P=4x+2(6-x^2)\). Before we consider graphing these to identify the max/min, we should seriously reflect on what a reasonable domain for the these function should be. Since \(x\) is the distance from the origin to the right, it makes sense that it should be greater than or equal to zero. However, the edge of the parabola to the right is the furthest we can go, so we should solve \(0=6-x^2\) to determine that right endpoint. Solving this gives \(x=\sqrt{6}\), where we take the positive root since the negative is to the left. Thus the domain is \([0,\sqrt{6}]\). When graphing (for instance on a TI device) we should set the window so that the minimum \(x\) value is 0 and the maximium \(x\) value is \(\sqrt{6}\).

First up, here is the graph of the area function:

And then the graph of the perimeter function:

We can use the calculator functions to determine the local min/max and the global min/max too.