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

Maximize light with width \(w\) and height \(h = \frac{P - w - \frac{\pi w}{2}}{2}\) under the given perimeter.

Step 1: Understand the Problem

We have a window that consists of a rectangle and a semicircle on top. The problem states that the total perimeter is fixed, and we want to optimize the light passing through by adjusting the proportions of the rectangle and the semicircle.

Step 2: Define Variables

Let the width of the rectangle be \( w \) and the height be \( h \). The radius of the semicircle will be \( r = \frac{w}{2} \). The total perimeter of the semicircle and rectangle is given by \( 2h + w + \pi r = P \), where \( P \) is the fixed perimeter.

Step 3: Express Light Transmission

The area of the rectangle is \( A_{rect} = w \times h \), and for the semicircle it is \( A_{semi} = \frac{1}{2} \pi \left(\frac{w}{2}\right)^2 \). Because tinted glass transmits half the light, effective area contribution becomes \( A = A_{rect} + \frac{1}{2} A_{semi} \).

Step 4: Optimize the Light

Substitute \( r = \frac{w}{2} \) into the perimeter equation to express \( h \) in terms of \( w \). Then substitute \( h(w) \) into the light transmission area function \( A(w) \). Find the derivative \( A'(w) \) and set it to zero to find critical points, which maximizes light.

Step 5: Solve Perimeter Constraint

From the perimeter constraint you have \( 2h + w + \pi (\frac{w}{2}) = P \). Rearrange it to find \( h = \frac{P - w - \frac{\pi w}{2}}{2} \).

Step 6: Substitute into Light Area Function

Substitute \( h(w) \) into the effective area formula: \( A(w) = w \cdot \left(\frac{P - w - \frac{\pi w}{2}}{2}\right) + \frac{1}{2} \cdot \frac{1}{2} \pi \left(\frac{w}{2}\right)^2 \). Simplify and take the derivative with respect to \( w \).

Step 7: Find Critical Points

After differentiating, set \( A'(w) = 0 \) to find possible values of \( w \). Solve for \( w \) and ensure it lies within realistic bounds.

Step 8: Conclusion on Proportions

After solving, you find the dimensions that maximize the light through the window by calculating \( h \) using the dimensions found for \( w \). Check this solution against the fixed perimeter constraint to verify feasibility.

Key Concepts

Light Transmission

The challenge with optimizing a window is managing how much light it lets through. Light transmission measures how much light gets through different materials. Clear glass lets in all the light that hits it. Tinted glass lets through only a fraction, in this case, half.

The window is partly made of clear glass (the rectangle) and partly of tinted glass (the semicircle). Optimizing light transmission means maximizing the area that most efficiently lets light through. The effective light area is not merely the sum of clear and tinted glass areas. Instead, it equals the clear glass area plus half of the tinted glass area.

This consideration ensures a comprehensive approach to maximizing the window's light transmission efficiency.

Perimeter Constraint

Constraints are limits placed on problems that can't be exceeded. For this window, the specified constraint is the total perimeter, which remains fixed.

Knowing the window consists of a rectangle and a semicircle, the perimeter is calculated with the formula:

• Total Perimeter = 2h + w + \( \pi \cdot r \)

Here, 'h' is the rectangle's height, 'w' is its width, and 'r' is the semicircle's radius, \(r = w/2\). P represents the fixed perimeter.

Solving for one dimension with this constraint lets you express other dimensions. For example, rearranging gives height 'h' as a function of width 'w'. This relationship is vital because it links to optimizing light transmission.

Rectangular and Circular Geometry

The geometry of the window involves both rectangular and circular components, key for defining structure and calculations.

The rectangle is straightforward, characterized by its width 'w' and height 'h'. The semicircle, however, requires understanding of circular geometry. Its radius is half the rectangle's width (\(r = w/2\)). The area of the semicircle is half of a full circle, \(A_{semi} = \frac{1}{2} \cdot \pi \cdot \left(\frac{w}{2}\right)^2\).

Comprehending these shapes helps us compute areas and, more importantly, transform the complex window unit into individual parts. Understanding how they fit together aids in analyzing and optimizing light transmission.

Critical Points in Calculus

Calculus helps us find optimal solutions using derivatives to pinpoint critical points. Critical points occur where the derivative of a function is zero or undefined. In this problem, it measures where light transmission is optimized based on the width 'w'.

By deriving the light transmission area function with respect to 'w', and setting the derivative to zero, we find these critical points. Solutions to this equation show us the width that maximizes light, respecting both the materials’ light transmission and the geometric constraint.

After solving for 'w', checking if these points represent maxima or minima ensures we choose the best solution. This process uses the power of calculus to provide a structured route to optimize designs.