Questions: A farmer has 300 ft of fencing and wants to enclose a rectangular area of 5000 ft^2. What dimensions should the farmer use? The length of the longer side of the fence is ft and the length of the shorter side of the fence is ft. (Simplify your answers.)

Solution Steps

To solve this problem, we need to set up a system of equations based on the given information. We know the perimeter of the rectangle and the area. We can use these equations to solve for the dimensions of the rectangle.

1. Let \( l \) be the length and \( w \) be the width of the rectangle.
2. The perimeter equation is \( 2l + 2w = 300 \).
3. The area equation is \( l \times w = 5000 \).

We can solve these equations simultaneously to find the values of \( l \) and \( w \).

We start with the equations based on the problem statement. The perimeter of the rectangle is given by:

\[ 2l + 2w = 300 \]

And the area is given by:

\[ lw = 5000 \]

From the perimeter equation, we can simplify it to:

\[ l + w = 150 \]

Now we can express \( w \) in terms of \( l \):

\[ w = 150 - l \]

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

\[ l(150 - l) = 5000 \]

Expanding this, we have:

\[ 150l - l^2 = 5000 \]

Rearranging leads to the quadratic equation:

\[ l^2 - 150l + 5000 = 0 \]

Using the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -150 \), and \( c = 5000 \):

\[ l = \frac{150 \pm \sqrt{(-150)^2 - 4 \cdot 1 \cdot 5000}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (-150)^2 - 4 \cdot 1 \cdot 5000 = 22500 - 20000 = 2500 \]

Thus, we find:

\[ l = \frac{150 \pm 50}{2} \]

This gives us two possible solutions for \( l \):

\[ l = 100 \quad \text{or} \quad l = 50 \]

Correspondingly, the values for \( w \) are:

\[ w = 50 \quad \text{or} \quad w = 100 \]

Final Answer