Bill wrote:
Lew Hodgett wrote:
Bill wrote:
A farmer has 20 feet of fencing and wishing to build a rectangular
pen against his barn (by adding 3 sides), maximizing the area. What
should the dimensions be? Note: It makes the kids smile if I
include a little animal in the diagram.
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Not enough information to develop 3 independent equations.

You can use Taylor's theorem of approximation to determine
that a 5 ft x 10 ft pen with the 10 ft side parallel to barn wall will
yield max area.
-----------------------------------------------------------------
W = Width
L = Length
A = Area

2W + L = 20

A = WL
-------------------------------------------------------------
WHEN W = 7, THEN L = 6
A = 42

WHEN W = 6, THEN L = 8
A = 48

WHEN W = 5, THEN L = 10
A = 50

WHEN W = 4, THEN L = 12
A = 48

WHEN W = 3, THEN L = 14
A = 42

Ye gads you are making me dig.

Haven't used most of this stuff in over 50 years.

Lew



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. ; )