Problem 13 - Calculus with Concepts in Calculus

Question

An outdoor track is to be created in the shape of a rectangle with semicircles at two opposite ends. If the perimeter of the track is 440 yards, find the dimensions of the track for which the area of rectangular portion is maximized.

Step-by-Step Solution

Answer

The area is maximized for a rectangle of length 110 yards and width \(\frac{220}{\pi}\) yards.

Step 1: Understanding the Problem

The problem involves a perimeter made up of a rectangular part and semicircles. The total perimeter is 440 yards. We need to find dimensions that maximize the area of the rectangle.

Step 2: Define Variables

Let the length of the rectangle be \( L \) and width (diameter of the semicircle) be \( 2R \). The semicircles form a complete circle with a circumference of \( 2\pi R \).

Step 3: Set Up Perimeter Equation

The perimeter equation is the sum of the perimeter of the rectangle and the circle: \[ 2L + 2\pi R = 440. \] Solve for \( L \) to express it in terms of \( R \): \[ L = 220 - \pi R. \]

Step 4: Express Area of the Rectangle

The area \( A \) of the rectangle is \( L \times 2R \): \[ A = (220 - \pi R)(2R). \] Simplify: \[ A = 440R - 2\pi R^2. \]

Step 5: Find Maximum Area Using Optimization

To find the value of \( R \) that maximizes \( A \), take the derivative of \( A \) with respect to \( R \): \[ \frac{dA}{dR} = 440 - 4\pi R. \] Set the derivative equal to zero and solve for \( R \):\[ 440 - 4\pi R = 0, \quad \Rightarrow \quad R = \frac{440}{4\pi}. \]

Step 6: Calculate the Length of the Rectangle

Substitute \( R = \frac{440}{4\pi} \) back into the equation for \( L \): \[ L = 220 - \pi \left(\frac{440}{4\pi}\right), \quad \Rightarrow \quad L = 220 - 110 = 110. \]

Step 7: Determine the Final Dimensions

So, the dimensions that maximize the area of the rectangular portion are: - Length \( L = 110 \) yards - Width (diameter) \( 2R = \frac{440}{2\pi} = \frac{220}{\pi} \) yards.

Key Concepts

Perimeter Problems

In calculus, perimeter problems are common scenarios where you analyze shapes to optimize certain characteristics, often looking for maximal or minimal possible values. In these situations, the question typically involves a fixed perimeter, and you need to find dimension measurements that satisfy additional conditions, such as maximizing area.

In our track example, the combined shape has a rectangular body with semicircles at either end. The total perimeter given is 440 yards, which includes the length of the rectangle and the distance around the semicircles. To start solving perimeter problems, always:

• Identify the different components contributing to the perimeter, like sides of rectangles or circumferences of circles formed by semicircles.
• Express the perimeter requirement in a mathematical equation.

Using these steps helps translate the physical problem into a solvable mathematical form.

Area Maximization

In scenarios where maximizing area is the goal, the objective is to determine dimensions that yield the largest possible space within a given shape. This often requires calculus techniques, particularly when dealing with auxiliary constraints like a fixed perimeter.

For area maximization to occur, one must:

• Define the function that represents the area, using variables that are related through known constraints like the perimeter.
• Use calculus methods, such as taking derivatives, to find critical points where the function attains maximum values.

In this exercise, the area of the rectangular part is represented by the equation: \[ A = 440R - 2\pi R^2 \]where \( R \) is related to the semicircles. Solving this with calculus ensures you find the dimensions that make the area as large as possible.

Geometry in Calculus

Geometry and calculus frequently intersect when you need to deal with shapes in optimization problems. Concepts from geometry help in establishing relationships within the given shapes and translating them into equations necessary for calculus optimization.

For instance, when addressing a shape like our track, it blends basic geometric shapes—a rectangle and a complete circle made from semicircles. This requires understanding:

• How each part of these shapes contributes to dimensions like perimeter and area.
• The relationships between geometric properties, such as how semicircle diameters relate to the full circle's circumference.
• Special equations, like the circumference of a circle given by \( C = 2\pi R \), which are foundational in setting up optimization problems.

Grasping these geometric ideas helps you mold real-world problems into a form ready to be tackled with calculus techniques.