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 x\displaystyle x meters.
The length (the side parallel to the river) is 40002x\displaystyle 4000-2x

Area = width X length

A(x)=x(40002x)\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:

x(40002x)=0\displaystyle x(4000-2x)=0

So our graph crosses the x-axis when x=0\displaystyle x = 0 (on its way up) and again when (40002x)=0\displaystyle (4000 - 2x) = 0 (on its way down). Solving the second factor, we get:

40002x=0\displaystyle 4000-2x=0

2x=4000\displaystyle -2x=-4000

x=2000\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: 0+20002=1000\displaystyle \frac{0+2000}{2}=1000 meters.

So the maximum area occurs when

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

The length = 40002x=400021000=2000\displaystyle 4000-2x = 4000-2 \cdot 1000= 2000 meters.

The maximum area is 20001000=2000000\displaystyle 2000 \cdot 1000 = 2000000 square meters.

 
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