A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter of the track is to be 1300 meters. What should the dimensions of the rectangle be so that the area of the rectangle is​ maximized?
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Part 1
The dimensions of the rectangle are

enter your response here meters.
​(Use a comma to separate answers as needed. Type an integer or decimal rounded to the nearest hundredth as​ needed.)

The given problem involves a track and field area in the shape of a rectangle with semicircles at each end. To find the dimensions of the rectangle that maximize the area, we need to use calculus and some geometric principles. Let's break it down.

### Step 1: Define variables
Let:
- \( L \) be the length of the rectangle (parallel to the straight sides of the track),
- \( W \) be the width of the rectangle (equal to the diameter of each semicircle).

The perimeter of the track includes:
- The two lengths of the rectangle, \( 2L \),
- The circumference of the two semicircles (which together form a full circle), \( \pi W \).

The total inside perimeter of the track is given as 1300 meters. So, the perimeter equation is:
\[
2L + \pi W = 1300
\]

### Step 2: Express \( L \) in terms of \( W \)
We can rearrange the perimeter equation to express \( L \) in terms of \( W \):
\[
L = \frac{1300 - \pi W}{2}
\]

### Step 3: Maximize the area
The area of the rectangle is:
\[
A = L \times W
\]

Substitute \( L \) from the equation above:
\[
A = \left( \frac{1300 - \pi W}{2} \right) W
\]
\[
A = \frac{1300W - \pi W^2}{2}
\]

### Step 4: Differentiate the area function
To maximize the area, we differentiate \( A \) with respect to \( W \) and set the derivative equal to zero:
\[
\frac{dA}{dW} = \frac{1300 - 2\pi W}{2}
\]
Set this equal to zero:
\[
1300 - 2\pi W = 0
\]
Solve for \( W \):
\[
W = \frac{1300}{2\pi} \approx 206.64 \, \text{meters}
\]

### Step 5: Find \( L \)
Now that we have \( W \), substitute it back into the expression for \( L \):
\[
L = \frac{1300 - \pi \times 206.64}{2} \approx 146.68 \, \text{meters}
\]

### Final Answer:
The dimensions of the rectangle are approximately \( 146.68 \) meters by \( 206.64 \) meters.

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How do we confirm that the area is maximized?
2. What if the perimeter was different? How would that change the dimensions?
3. Can this process be generalized for any shape with semicircular ends?
4. What role does the derivative play in maximizing functions?
5. How can the use of Lagrange multipliers apply to this type of problem?

**Tip:** Always check if the second derivative is negative to confirm that you have a maximum in optimization problems.

A track and field playing area is in the shape of a rectangle with semicircles at each end. The inside perimeter of the track is to be 1300 meters. What should the dimensions of the rectangle be so that the area of the rectangle is maximized? The dimensions of the rectangle are ___ meters. (Use a comma to separate answers as needed. Type an integer or decimal rounded to the nearest hundredth as needed.)

This problem focuses on finding the dimensions of a rectangular track with semicircular ends that maximizes the area while keeping the inside perimeter at 1300 meters. Learn how to use calculus and geometry principles to solve this optimization problem. Suitable for high school students in Grades 10-12.

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