A running track of 440 ft is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then find the length of its sides.

Answer

To solve the problem of maximizing the area of a rectangular football field enclosed by a running track, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = length of the rectangular portion
- \( y \) = width of the rectangular portion

### Step 2: Understand the Track Layout
The track consists of:
- Two lengths of the rectangle (each of length \( x \))
- Two semicircles at each end (which together form a full circle with diameter \( y \))

### Step 3: Write the Perimeter Equation
The total perimeter of the running track is given as 440 ft. The perimeter can be expressed as:
\[ P = 2x + \pi y \]
Since the semicircles together form a full circle, we can use the circumference formula for the circle.

### Step 4: Set Up the Equation
We can set up the equation based on the perimeter:
\[ 2x + \pi y = 440 \]

### Step 5: Solve for \( y \)
Rearranging the equation to express \( y \) in terms of \( x \):
\[ \pi y = 440 - 2x \]
\[ y = \frac{440 - 2x}{\pi} \]

### Step 6: Write the Area Function
The area \( A \) of the rectangle is given by:
\[ A = x \cdot y \]
Substituting the expression for \( y \):
\[ A = x \cdot \frac{440 - 2x}{\pi} \]
\[ A = \frac{440x - 2x^2}{\pi} \]

### Step 7: Differentiate the Area Function
To find the maximum area, we differentiate \( A \) with respect to \( x \):
\[ \frac{dA}{dx} = \frac{440 - 4x}{\pi} \]

### Step 8: Set the Derivative to Zero
Setting the derivative equal to zero to find critical points:
\[ \frac{440 - 4x}{\pi} = 0 \]
\[ 440 - 4x = 0 \]
\[ 4x = 440 \]
\[ x = 110 \]

### Step 9: Find \( y \)
Substituting \( x = 110 \) back into the equation for \( y \):
\[ y = \frac{440 - 2(110)}{\pi} \]
\[ y = \frac{440 - 220}{\pi} \]
\[ y = \frac{220}{\pi} \approx 70 \]

### Step 10: Conclusion
Thus, the dimensions of the rectangular portion that maximize the area are:
- Length \( x = 110 \) ft
- Width \( y \approx 70 \) ft