Bill wrote:

Instead of using 2 variables L and W, try it with just 1 (say just
L or W), you can write the area as a single equation/function of
it. You'll have a quadratic equation whose graph is a parabola...
This will lead you not only an answer, but the fact that the answer
lies at the vertex of a parabola opening downwards will justify for
you that it is the unique best answer.


Here is a start.
Let L denote the length. Then the width equals (20-L)/2.

So the area = L(20-L)/2=-.5L^2 +10L, a quadratic equation in terms
of the single variable L.

You may find the vertex of the graph of A = -.5L^2 +10L in several
ways (as a local max in calculus, completing the square, using it's
symmetry about it's axis of symmetry--which is the midpoint of it's
L-intercepts, for instance (these can be found with the quadratic
formula or just by factoring)).

Any of these will show you why the answer you already found really
is the correct and
only one. ; )
-----------------------------------------------
Too many years and too many beers.

Thanks for the input.

Lew