Enclosing the MOst Area with a fence

karliekay

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Sep 21, 2010
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A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed.


it states that the width is x and the length is 4000-2x.

I know that the Perimeter is 4000. So, 4000=L+2w because we are not using on of the lengths.

What I did is inputed 4000-2x in for L, but I got 0. What do I do?
 
karliekay said:
A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed.


it states that the width is x and the length is 4000-2x.

I know that the Perimeter is 4000. So, 4000=L+2w because we are not using on of the lengths.

What I did is inputed 4000-2x in for L, but I got 0. What do I do?
Hi karliekay,

You have a rectangle with one side being the river.

We know the width (the sides perpendicular to the river) to be \(\displaystyle x\) meters.
The length (the side parallel to the river) is \(\displaystyle 4000-2x\)

Area = width X length

\(\displaystyle A(x) = x(4000-2x)\)

The graph of this quadratic function is a parabola. Where the graph crosses the x-axis, the value of the area (A) must be zero. Therefore, our equation becomes:

\(\displaystyle x(4000-2x)=0\)

So our graph crosses the x-axis when \(\displaystyle x = 0\) (on its way up) and again when \(\displaystyle (4000 - 2x) = 0\) (on its way down). Solving the second factor, we get:

\(\displaystyle 4000-2x=0\)

\(\displaystyle -2x=-4000\)

\(\displaystyle x=2000\)

This tells you that the graph crosses the x-axis at 0 and 2000. Since a parabola is symmetrical, the x-coordinate of the vertex will be halfway between these two values: \(\displaystyle \frac{0+2000}{2}=1000\) meters.

So the maximum area occurs when

The width = \(\displaystyle x = 1000\) meters.

The length = \(\displaystyle 4000-2x = 4000-2 \cdot 1000= 2000\) meters.

The maximum area is \(\displaystyle 2000 \cdot 1000 = 2000000\) square meters.

 
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