Jonathan W. answered 05/18/15
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Is this for calculus? If not, you can draw various rectangles with area 72 (1 × 72, 2 × 36, ... , 8 × 9, maybe even √72 × √72), and compare their perimeters (2·1 + 2·72, ...). Your answer for the smallest should match your intuition.
If this is for calculus, you can verify your intuition by calculating the dimensions. Call the length and width x and y. Then the area xy = 72, and the perimeter 2x + 2y is to be optimized. To make the perimeter a function of one variable, substitute y = 72/x from the area constraint in for y. Then the perimeter is
P(x) = 2x + 2(72/x)
Now take the derivative of P(x) and set it to 0. (This is where the graph of P(x) levels off.)
P'(x) = 0
Solve for x, plug that into P(x), and you'll get the smallest perimeter. As for the largest perimeter, there is none.