Suppose you have 40 meters of fencing what is the greatest rectangular area that you can enclose

It'd help if you could sketch this situation.  Note that the area of a rectangle is equal to the product of its width and length:  A = L W.

Consider the perimeter of this rectangular area.  It's P = 2 L + 2 W.  Note that P = 40 meters in this problem.

Thus, if we choose to use W as our independent variable, then P = 40 meters = 2 L + 2 W.  Let's express L in terms of W.  Divide both sides of the following equation by 2:  40 = 2 L + 2 W.  We get 20 = L + W.  Thus, L = 20 - w.

Then the area of the rectangle is  A = ( 20 - W)*W.

Multiply this out.  Your result will be a quadratic equation.  Graph this quadratic equation (in other words, graph the function that represents the area of the rectangle).  For which W value is the area at its maximum?  

Alternatively, find the vertex of this graph:  remember that the x- (or W-) coordinate  of the vertex is given by

W = -b/(2a), where a is the coefficient of W^2 an b is the coefficient of W in your quadratic equation.

Finally, substitute this value of W into your quadratic equation, to calculate the maximum area.