Problem 351 - Calculus Volume 1

Question

Set up, but do not evaluate, each optimization problem. A window is composed of a semicircle placed on top of a rectangle. If you have \(20 \mathrm{ft}\) of window-framing materials for the outer frame, what is the maximum size of the window you can create? Use \(r\) to represent the radius of the semicircle.

Step-by-Step Solution

Answer

Maximize: \( A = 2r \left( \frac{20 - 2r - \pi r}{2} \right) + \frac{1}{2}\pi r^2 \), constrained by \( 2h + 2r + \pi r = 20 \).

Step 1: Understand the Geometry

The window consists of a rectangle with a semicircle on top. The semicircle's diameter is the same as the width of the rectangle, which is \( 2r \); therefore, the radius is \( r \).

Step 2: Define Variables

Let \( r \) be the radius of the semicircle, and\( h \) be the height of the rectangle. The total perimeter of the window (semicircle + rectangle) needs to be equal to the available 20 ft of framing material.

Step 3: Identify the Perimeter Constraints

The perimeter consists of the rectangle's two heights and width plus the semicircle's circumference. Thus, \[ P = 2h + 2r + \pi r = 20 \] since the semicircle's circumference is \( \pi r \) and the rectangle's width is \( 2r \).

Step 4: Formulate the Objective Function

We want to maximize the total area \( A \) of the window: \[ A = \text{Area of rectangle} + \text{Area of semicircle} = 2rh + \frac{1}{2}\pi r^2 \].

Step 5: Setup the Optimization Problem

Use the constraint \( 2h + 2r + \pi r = 20 \) to express \( h \) in terms of \( r \):\[ h = \frac{20 - 2r - \pi r}{2} \]. Substitute \( h \) in the objective function \( A \):\[ A = 2r \left( \frac{20 - 2r - \pi r}{2} \right) + \frac{1}{2}\pi r^2 \]. This is the function we need to maximize.

Key Concepts

Semicircle

In this optimization problem, the semicircle is a significant part of the window shape. A semicircle is half of a circle, therefore it occupies a distinct space atop the rectangle to form the window structure. The semicircle's defining feature is its radius, denoted as \( r \), and its diameter, which is twice the radius, \( 2r \). This diameter also serves as the width of the rectangle beneath it.
What's crucial here is the perimeter consideration. The perimeter contribution of the semicircle is half of the full circle's circumference, given by the formula \( \pi r \), where \( \pi \) is approximately 3.14159. This understanding of the semicircle's dimensions helps in forming the perimeter constraint that combines the rectangle and semicircle into a single, cohesive structure.

Rectangle

A rectangle is a fundamental geometric shape, and in this problem, it forms the lower portion of the window. The rectangle's width is equal to the diameter of the semicircle, which means it is \( 2r \). The height of the rectangle is a separate variable, \( h \), which we will determine using the perimeter constraints.
The problem requires maximizing the window's area, which includes both the rectangle and the semicircle. For a rectangle, the area is calculated by multiplying the width and the height, expressed as \( 2rh \). Understanding the rectangle's relationship with the semicircle is essential, as both parts share a common width, and their dimensions influence each other through the constraints provided.

Perimeter Constraints

Perimeter constraints play a pivotal role in optimization problems, as they limit the size of the object created. In this exercise, we're limited to 20 feet of framing material for the entire outer frame of the window, which includes both the semicircle and rectangle.
The formula expressing this constraint is \( 2h + 2r + \pi r = 20 \), which accounts for two heights and the width from the rectangle, as well as the semicircle's half-circumference. This equation is vital because it ties together the elements of the window, enabling us to solve for one variable in terms of another, and eventually substitute that into our objective function. Properly understanding and applying perimeter constraints is essential to solving the problem effectively.

Objective Function

The objective function is the mathematical expression we aim to maximize or minimize in optimization problems. Here, our goal is to maximize the total area \( A \) of the window, which is composed of two parts: the area of the rectangle and the area of the semicircle.
The area of the rectangle is \( 2rh \), while the area of the semicircle is \( \frac{1}{2} \pi r^2 \). Combining these gives the total area as \( A = 2rh + \frac{1}{2} \pi r^2 \). To solve this optimization problem, we need to express \( h \) in terms of \( r \) using the perimeter constraint, leading to the function being dependent solely on \( r \).Substituting \( h \) from the perimeter constraint \[ h = \frac{20 - 2r - \pi r}{2} \]into our objective function, we get a new expression for \( A \) that depends only on \( r \). This enables us to apply calculus techniques to find the value of \( r \) that maximizes the window's area.