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**Suppose a rancher wants to enclose an area along the side
an existing fence. If the enclosed area
is to be rectangular and the rancher has a fixed amount of fencing material to
make the other three sides, what ratio of length to width will result in the
largest possible area? **

**[Problem submitted by Vin Lee, LACC Associate Professor
of Mathematics.]**

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**Solution:**

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**Let p be the total length of the 3 new sides of the
rectangle, x the length of the side parallel to the old fence, and y the length
of each of the 2 sides perpendicular to the old fence. Then **

**
_{} **

** _{}**

** Area
= xy**

** _{}**

** _{}**

**This is a quadratic equation in standard form. If it is graphed with Area on the vertical
axis and y on the horizontal axis, then the graph is a downward opening
parabola whose maximum value of Area occurs at the vertex when _{}. So, **

** _{}**

** _{}**

** _{}.**

**So, the maximum area occurs when the ratio of length to
width is _{} = 2. _{}**