Problem 22 - Thomas Calculus in SI Units

Question

A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

Step-by-Step Solution

Answer

Optimal proportions are where \( w = \frac{P}{2(1 + \frac{\pi}{4})} \) and corresponding \( h \).

Step 1: Introduce the Variables

Let the width of the rectangular part of the window be \( w \), and the height be \( h \). The semicircle sits on top of the rectangle, having a diameter equal to \( w \), so its radius is \( r = \frac{w}{2} \).

Step 2: Write the Perimeter Equation

The total perimeter of the window involves two heights \( h \), one width \( w \), and the perimeter of the semicircle \( \pi \frac{w}{2} \). Thus, the perimeter equation is given by:\[ P = 2h + w + \pi \frac{w}{2} \]

Step 3: Express Light Transmission

Assuming unit areas transmit light, the rectangle transmits \( w \times h \) and the semicircle, which transmits half as much light, transmits \( \frac{1}{2} \times \frac{\pi}{2} \left(\frac{w}{2}\right)^2 = \frac{\pi w^2}{8} \). The total light transmitted is:\[ L = wh + \frac{\pi w^2}{8} \]

Step 4: Solve the Perimeter Constraint for One Variable

Solve the perimeter equation for \( h \):\[ h = \frac{P - w - \pi \frac{w}{2}}{2} \]

Step 5: Substitute and Simplify Light Equation

Substitute \( h \) from the perimeter equation into the light equation:\[ L = w \left(\frac{P - w - \pi \frac{w}{2}}{2}\right) + \frac{\pi w^2}{8} \]Simplify to:\[ L = \frac{w(P - w - \pi \frac{w}{2})}{2} + \frac{\pi w^2}{8} \]

Step 6: Take Derivative of Light Function

Take the derivative \( \frac{dL}{dw} \) to find the critical points:\[ \frac{dL}{dw} = \frac{P}{2} - w - \frac{\pi w}{4} + \frac{\pi w}{4} \]Simplify to:\[ \frac{dL}{dw} = \frac{P}{2} - w - \frac{\pi w}{4} \]

Step 7: Set Derivative to Zero and Solve

Set \( \frac{dL}{dw} = 0 \) and solve for \( w \):\[ \frac{P}{2} - w - \frac{\pi w}{4} = 0 \]This simplifies further to:\[ w(1 + \frac{\pi}{4}) = \frac{P}{2} \]So,\[ w = \frac{P}{2(1 + \frac{\pi}{4})} \]

Step 8: Calculate Heights and Verify Solutions

Use \( w \) to find \( h \) using \[ h = \frac{P - w - \pi \frac{w}{2}}{2} \]. Verify solution by ensuring all perimeter lengths are correct. Check the second derivative to ensure a maximum is found.

Key Concepts

Perimeter constraints in optimization problems

In designing a geometric shape, such as the window in this problem, perimeter constraints are crucial. The perimeter determines the total length around the shape. Understanding and applying these constraints help in optimizing various properties like area or light transmission.
For this window, the perimeter includes both the rectangle and the semicircle. It's expressed as:

• The sum of the height of two sides of the rectangle (2h).
• The width at the base of the rectangle (w).
• The curved path across the top, which is the semicircle's perimeter \( \pi \frac{w}{2} \).

By writing all these together, we get the equation: \[ P = 2h + w + \pi \frac{w}{2} \]
This equation allows us to express one geometric variable in terms of others. In optimization, you solve these types of equations to either minimize or maximize certain features, like maximizing light transmission in this instance.

Balancing light transmission with material constraints

Designing windows often requires balancing light transmission with structural constraints, like perimeter or area limitations. In this scenario, the window is made of two parts: a rectangle of clear glass and a semicircle of tinted glass.
The clear rectangle naturally allows more light through. Thus, its light transmission is modeled simply as its area (width x height). The tinted semicircle transmits less light per unit area due to its material property, capturing only half the light, modeled as \( \frac{\pi w^2}{8} \).
Formulating light transmitted as a function of these areas helps in understanding how changing dimensions – constrained by total perimeter – affects light intake. The equation:

• \( L = wh + \frac{\pi w^2}{8} \)

helps us focus adjustments on dimensions like height and width to optimize for maximum light.

Applying calculus derivatives to find optimal dimensions

Calculus gives us powerful tools to find maximum and minimum values – often pivotal in optimization problems. Here, derivatives allow us to determine specific dimensions where the light transmission is optimized under given constraints.
The primary aim is to find the critical points where the rate of change of light (with respect to width, in this case) becomes zero, indicating a possible maximum light intake.
First, the derivative of the light function \( \frac{dL}{dw} \) is calculated:

• Derivative: \( \frac{P}{2} - w - \frac{\pi w}{4} \).

Setting this to zero lets us solve for the optimal width (w), ensuring maximum light admission within the perimeter limit. Once found, these values help compute the height using the constraint, ensuring all conditions are satisfied as per the equation.