Problem 22 - Thomas Calculus

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

To maximize light, choose window dimensions satisfying the derivative of the light function based on perimeter constraint.

Step 1: Define the Variables

Let the width of the rectangle be \( w \) and the height be \( h \). Since the semicircle is on top of the rectangle, its diameter is equal to the width \( w \). The radius of the semicircle is therefore \( \frac{w}{2} \). This means that the total height of the window is \( h + \frac{w}{2} \).

Step 2: Express the Perimeter Constraint

The total perimeter of the window includes the two heights of the rectangle, the width of the rectangle, and the perimeter of the semicircle. Since the circumference of a full circle is \( \pi d \), the perimeter of the semicircle is \( \frac{\pi w}{2} \). Therefore, the perimeter constraint is given by: \[ 2h + w + \frac{\pi w}{2} = P \] where \( P \) is the fixed total perimeter.

Step 3: Light Transmission Formulation

Let \( A_r = w \cdot h \) be the area of the rectangle, and let \( A_s = \frac{1}{2}\pi\left(\frac{w}{2}\right)^2 = \frac{\pi w^2}{8} \) be the area of the semicircle. The light admitted by the rectangle is proportional to \( A_r \) and the light admitted by the semicircle is proportional to \( \frac{A_s}{2} \) since it is tinted and transmits half as much light. Therefore, the total light transmitted is: \[ L = A_r + \frac{A_s}{2} = wh + \frac{\pi w^2}{16} \].

Step 4: Substituting the Perimeter Constraint

To maximize the light, substitute the perimeter constraint in terms of one variable. Solve for \( h \) from: \[ 2h + w + \frac{\pi w}{2} = P \]\[ h = \frac{P - w - \frac{\pi w}{2}}{2} \]Substitute \( h \) in the light expression: \[ L = w\left(\frac{P - w - \frac{\pi w}{2}}{2}\right) + \frac{\pi w^2}{16} \].

Step 5: Maximize the Light Function

To find the maximum light, differentiate \( L \) with respect to \( w \), set the derivative equal to zero, and solve for \( w \). This involves finding:\[ \frac{dL}{dw} = \frac{P - w - \frac{\pi w}{2}}{2} - \frac{w}{2} - \frac{\pi w}{4} + \frac{\pi w}{8} = 0 \].

Step 6: Solve the Derivative Equation for \( w \)

Solving the above equation gives: \[ P - w - \frac{\pi w}{2} = \frac{w + \pi w}{4} \]Solve for \( w \) and substitute back to get \( h \).Finally, determine the relations for maximum light based on the solved values.

Key Concepts

Perimeter constraints

Understanding perimeter constraints is crucial when optimizing shapes to achieve specific goals. In this exercise, the perimeter constraint refers to the fixed outline or boundary of the window. This boundary encompasses the entire edge of the window, which consists of both the rectangle and the semicircle. Let’s break it down:

• The rectangle requires two heights and one width, accounting for its sides, contributing to the perimeter.
• The semicircle, which tops the rectangle, adds additional perimeter based on its curved edge. A full circle's circumference is calculated using the formula \( \pi d \), where \( d \) is the diameter. The semicircle’s perimeter is thus half of that.

Together, these elements form the perimeter constraint equation: \( 2h + w + \frac{\pi w}{2} = P \). The perimeter is fixed, meaning it does not change regardless of the dimensions of the rectangle and semicircle. By expressing all parts of the window in terms of simpler variables, we can find optimal dimensions that satisfy this constraint.

Light transmission

Light transmission refers to the amount of light passing through different parts of a material. In this problem, we need to maximize the light coming through a window made of clear and tinted glass.

• The rectangular part of the window is made of clear glass, allowing maximum light transmission according to its area \( A_r = wh \).
• The semicircle's glass is tinted, allowing only half as much light to pass through compared to the rectangular part. Thus, its light contribution is calculated as \( \frac{A_s}{2} = \frac{\pi w^2}{16} \), where \( A_s \) is the semicircle’s area.

To determine the optimal light transmission through the window, we combine the light contributions from both sections expressed in the formula: \( L = wh + \frac{\pi w^2}{16} \). This combined expression allows optimization by modifying the dimensions within the perimeter constraint.

Differentiation in calculus

Differentiation in calculus is a powerful tool to find where a function reaches its maximum or minimum value. Here, we use it to identify the window proportions that let in the most light.

• The goal is to maximize the light expression \( L = wh + \frac{\pi w^2}{16} \). We achieve this by expressing \( L \) in terms of a single variable, using the perimeter constraint equation.
• The perimeter constraint, when rearranged, provides \( h \) in terms of \( w \): \( h = \frac{P - w - \frac{\pi w}{2}}{2} \).

Now, substitute \( h \) back into the light expression to transform it entirely in terms of \( w \). Differentiating \( L \) with respect to \( w \) and setting the derivative to zero allows us to find critical values, indicating potential maximums or minimums:\[ \frac{dL}{dw} = \frac{P - w - \frac{\pi w}{2}}{2} - \frac{w}{2} - \frac{\pi w}{4} + \frac{\pi w}{8} = 0 \]Solving this derivative equation optimizes the window dimensions, ensuring the maximum amount of light is admitted given the perimeter constraint.