Problem 25 - Essential Calculus Early Transcendentals

Question

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30 \(\mathrm{ft}\) , find the dimensions of the window so that the greatest possible amount of light is admitted.

Step-by-Step Solution

Answer

The optimal width (\(x\)) and height (\(y\)) of the rectangle maximize the window's light admission.

Step 1: Define the Variables

Let the width of the rectangle be \( x \) and the height of the rectangle be \( y \). The radius of the semicircle, being half the diameter, is \( \frac{x}{2} \). The total perimeter of the window, including the curved part, needs to be considered.

Step 2: Expression for Perimeter

The perimeter of a Norman window is the sum of the perimeter of the rectangle and the arc of the semicircle. Therefore, the perimeter can be expressed as \( x + 2y + \frac{1}{2}\pi x = 30 \).

Step 3: Solve for One Variable

Solve the perimeter equation for \( y \):\[ y = \frac{30 - x - \frac{1}{2}\pi x}{2} \].

Step 4: Write Area Function

The area of the Norman window is the sum of the area of the rectangle \( xy \) and the semicircle \( \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 \). So, the area \( A \) is:\[ A = xy + \frac{1}{8}\pi x^2 \].

Step 5: Substitute for y in the Area Function

Substitute the expression for \( y \) from Step 3 into the area function:\[ A = x \left(\frac{30 - x - \frac{1}{2}\pi x}{2}\right) + \frac{1}{8}\pi x^2 \].

Step 6: Simplify the Area Function

Expand and simplify the expression obtained in Step 5 to get:\[ A = \frac{30x - x^2 - \frac{1}{2}\pi x^2}{2} + \frac{1}{8}\pi x^2 \].

Step 7: Further Simplification for Optimization

Combine like terms and simplify the expression further to get a quadratic function in terms of \( x \):\[ A = 15x - \left(\frac{1}{2} + \frac{1}{8}\pi\right)x^2 \].

Step 8: Find the Maximum Area using Derivative

To find the value of \( x \) that maximizes area, calculate the derivative of \( A \) with respect to \( x \) and set it to zero. Solve:\[ \frac{dA}{dx} = 15 - 2\left(\frac{1}{2} + \frac{1}{8}\pi\right)x = 0 \].

Step 9: Solve the Derivative Equation

Simplify and solve the derivative equation for \( x \):\[ x = \frac{15}{1 + \frac{1}{4}\pi} \].

Step 10: Calculate Corresponding y Value

Once \( x \) is found, calculate \( y \) using the equation from Step 3:\[ y = \frac{30 - x - \frac{1}{2}\pi x}{2} \].

Key Concepts

Perimeter Equation

In this optimization problem, we focus on the Norman window's perimeter. The perimeter includes the rectangle's sides and the semicircle's arc, relating to the window's overall shape. To write the perimeter equation, understand the components involved:

• Rectangle's width: Represented by \( x \)
• Rectangle's height: Represented by \( y \)
• Semicircle's diameter: Same as rectangle's width, \( x \), with the radius being \( \frac{x}{2} \)

The total perimeter, therefore, is the sum of these dimensions: two sides of width, two sides of height, and the semicircle's arc (which equals \( \frac{1}{2}\pi x \)). The equation becomes:\[x + 2y + \frac{1}{2}\pi x = 30\]This equation forms the basis for finding specific variable relationships needed later in the process.

Area Function

For the Norman window, we want to maximize the area through specific variable dimensions. The area includes both the rectangle's area and the semicircle's area. We express this through:

• Rectangle's area: \( xy \)
• Semicircle's area: \( \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 = \frac{1}{8}\pi x^2 \)

Combining these, the total area function \( A \) is:\[A = xy + \frac{1}{8}\pi x^2\]This function represents the complete expression for the area of the window, integrating both the lengths across the rectangle and the semicircle. Once we express \( y \) in terms of \( x \) using the perimeter equation, it allows a conversion of the area function to just depend on \( x \).

Quadratic Function

To find the optimal dimensions that maximize the area of the Norman window, we need an equation expressed as a quadratic function. After substituting \( y \) from the perimeter equation into our area function, we simplify it:\[A = \frac{30x - x^2 - \frac{1}{2}\pi x^2}{2} + \frac{1}{8}\pi x^2\]Combining like terms and simplifying gives us a quadratic function in terms of \( x \):\[A = 15x - \left(\frac{1}{2} + \frac{1}{8}\pi\right)x^2\]This quadratic function is crucial because solving it will pinpoint the \( x \)-value that maximizes the area. Such quadratic expressions typically involve a focused analysis on determining the vertex since it provides the extremum (maximum or minimum). In our case, we seek the maximum.

Derivative

Derivatives are a critical tool for finding maxima and minima of functions, especially relevant here for optimization. Taking a derivative of the area function \( A \), and setting it to zero, helps us find the value that gives the maximum area.The derivative of the quadratic area function in terms of \( x \) is:\[\frac{dA}{dx} = 15 - 2\left(\frac{1}{2} + \frac{1}{8}\pi\right)x\]Setting this derivative equal to zero allows us to solve for \( x \):\[15 - 2\left(\frac{1}{2} + \frac{1}{8}\pi\right)x = 0\]Upon solving, \( x \) is found using:\[x = \frac{15}{1 + \frac{1}{4}\pi}\]This provides the width of the rectangle that maximizes the area, with calculus' role being indispensable in this aspect to ensure the solution is accurate and efficient.