Problem 20 - Calculus: Early Transcendentals

Question

A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top. Suppose that the colored glass transmits only \(k\) times as much light per unit area as the clear glass \((k\) is between 0 and 1\() .\) If the distance from top to bottom (across both the rectangle and the semicircle) is a fixed distance \(H,\) find \((\) in terms of \(k)\) the ratio of vertical side to horizontal side of the rectangle for which the window lets through the most light.

Step-by-Step Solution

Answer

The optimal ratio of vertical to horizontal side is \( \frac{2 + k\pi}{4} \).

Step 1: Define Variables and Objective

Let the width of the rectangle be \(w\) and its height be \(h\). Hence, the radius of the semicircle is \(\frac{w}{2}\) because its diameter is equal to the width of the rectangle. The total height from the bottom of the rectangle to the top of the semicircle is \(H\), so \(h + \frac{w}{2} = H\). We want to maximize the light transmitted through the window.

Step 2: Calculate Total Light Transmission

The area of the rectangular glass is \(A_{\text{rect}} = w \cdot h\), and for the semicircle, it's \(A_{\text{semi}} = \frac{\pi \left(\frac{w}{2}\right)^2}{2} = \frac{\pi w^2}{8}\). The light transmitted through the rectangle is proportional to \(A_{\text{rect}}\), and through the semicircle is \(k \cdot A_{\text{semi}}\), because it transmits \(k\) times the light. Total light transmitted is proportional to \(w \cdot h + k \cdot \frac{\pi w^2}{8}\).

Step 3: Expressing Rectangle Dimensions in Terms of H

From \(h + \frac{w}{2} = H\), we can express \(h\) in terms of \(w\): \(h = H - \frac{w}{2}\). Substitute \(h\) in the expression for total light transmission: \(L = w(H - \frac{w}{2}) + k \cdot \frac{\pi w^2}{8}\).

Step 4: Derivative and Optimize Light Transmission

To find the optimum \(w\) for maximum light, take the derivative of \(L\) with respect to \(w\) and set it to zero: \(\frac{dL}{dw} = H - w + k \cdot \frac{\pi w}{4} = 0\). Solving for \(w\), we have: \[w = \frac{4H}{4 + k\pi}\].

Step 5: Calculate Ratio of Vertical to Horizontal Side

Substitute \(w\) back into the expression for \(h\), which is \(h = H - \frac{w}{2} = H - \frac{2H}{4 + k\pi} = \frac{H(2 + k\pi)}{4 + k\pi}\). Hence the ratio of vertical side to horizontal side of the rectangle is:\[ \frac{h}{w} = \frac{\frac{H(2 + k\pi)}{4 + k\pi}}{\frac{4H}{4 + k\pi}} = \frac{2 + k\pi}{4}\].

Key Concepts

Area Optimization

In any optimization problem, the goal is to find the best solution under a given set of constraints. For area optimization, we're tasked with either maximizing or minimizing the area involved in our problem. In this exercise, we're dealing with a window that consists of a rectangular section and a semicircular section.
The challenge is to optimize the light transmitted through the window, which involves maximizing the effective area through which light passes. The interesting twist is that the semicircular glass transmits less light, meaning any plan has to take these transmission rates into account.
By expressing the areas algebraically and considering the light transmittance factors, we set up an equation that represents the total light passing through the entire window. This equation will lead us towards understanding the optimal dimensions that allow for the maximum light to pass.

Derivative Applications

The derivative is a fundamental tool in calculus used to determine rates of change and to find optimal values, such as maxima or minima of functions. In this geometric problem, the concept of derivatives is applied to find the optimal width of the rectangle to maximize light transmission through the window.
Once we write an equation for the total light transmission, expressed in terms of the rectangle's width and height, we take the derivative of this equation with respect to the width. This helps us find the rate at which light transmission changes with respect to small changes in the width.
Setting the derivative equal to zero allows us to find critical points, which are potential candidates for maximum or minimum values. Through evaluating this critical point, we determine the optimal width that provides the desired maximum light transmission.

Geometric Problem-Solving

Geometric problem-solving involves a blend of algebraic manipulation and spatial reasoning. In solving for the optimized dimensions of the window, we encounter a series of geometric relations and constraints.
First, we note that the overall height from bottom to top, which involves both the rectangle and the semicircle, is fixed at height \( H \). This means that the height of the rectangle plus the effective radius of the semicircle equals \( H \).
Next, understanding the spatial layout and dimensions allows us to relate the height of the rectangle and the radius of the semicircle in a meaningful way. This highlights the importance of geometric understanding when setting up our problem. By exploring how different parameters relate and affect the total configuration, we systematically solve for the optimal geometric design of the window.

• Identifying fixed and variable dimensions.
• Understanding the relationship between dimensions and light transmission.
• Utilizing constraints to reach a practical solution.