A farmer is fencing off an area in his yard. He buys 50 feet of fencing to make a rectangular enclosure. He wants to make the fenced area as big as possible for the amount of fencing he has. In terms of the context, what is the maximum area he can create with the fencing?

Answer 1

To find the maximum area that the farmer can enclose using 50 feet of fencing in a rectangular shape, we can use some principles from geometry.

Let the length of the rectangle be \( l \) and the width be \( w \). The perimeter of the rectangle, which is the total length of the fencing, is given by the formula:

\[ P = 2l + 2w \]

Since the farmer has 50 feet of fencing, we set this equal to 50:

\[ 2l + 2w = 50 \]

We can simplify this equation:

\[ l + w = 25 \]

Now, the area \( A \) of the rectangle is given by:

\[ A = l \cdot w \]

To express the area in terms of one variable, we can solve for \( w \) in terms of \( l \):

\[ w = 25 - l \]

Substituting this expression for \( w \) into the area formula gives:

\[ A = l(25 - l) = 25l - l^2 \]

This is a quadratic equation in the standard form:

\[ A = -l^2 + 25l \]

The graph of this quadratic function is a downward-opening parabola. The maximum area occurs at the vertex of the parabola. The vertex \( l \) can be found using the formula:

\[ l = -\frac{b}{2a} \]

where \( a = -1 \) and \( b = 25 \):

\[ l = -\frac{25}{2 \cdot -1} = \frac{25}{2} = 12.5 \text{ feet} \]

Now substituting back to find \( w \):

\[ w = 25 - l = 25 - 12.5 = 12.5 \text{ feet} \]

So, both the length and the width of the rectangle that give the maximum area are each 12.5 feet. Now we can calculate the maximum area:

\[ A = l \cdot w = 12.5 \cdot 12.5 = 156.25 \text{ square feet} \]

Thus, the maximum area the farmer can create with 50 feet of fencing is

\[ \boxed{156.25} \text{ square feet}. \]