A window of perimeter P (including the base of the arch) is in the form of a rectangle surmounted by

Q: Wait, why do we include the partition line in the perimeter? Isn't the perimeter just the outer boundary?

That is a classic trap! In many geometry problems, 'perimeter' refers to the outer boundary, but here the problem explicitly states 'including the base of the arch'. This means the line segment separating the rectangle from the semi-circle is part of the material boundary that consumes the perimeter $P$. Always read the problem statement for these specific constraints.

Q: Why is the light transmission function $L = 3xy + \frac{\pi x^2}{8}$?

Great question. We are told the clear glass (the rectangle) transmits three times as much light per square meter as the colored glass (the semi-circle). If we assign a transmission factor of $1$ to the colored glass, the clear glass has a factor of $3$. Thus, $L = 3 \cdot \text{Area}_{\text{rect}} + 1 \cdot \text{Area}_{\text{semi}}$. Substituting the areas $xy$ and $\frac{\pi x^2}{8}$ gives you that exact expression.

Q: How do I know if the value of $x$ I found gives a maximum and not a minimum?

You can use the second derivative test. If you find $\frac{d^2L}{dx^2}$ and it is negative at your critical point, the function is concave down, confirming a maximum. Alternatively, since $L(x)$ is a downward-opening parabola (the coefficient of $x^2$ is negative), the vertex is guaranteed to be a maximum.

Question

A window of perimeter $P$ (including the base of the arch) is in the form of a rectangle surmounted by a semi circle. The semi- circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light?

Sigma Percentile · Accepted Answer

### The Architect's Dilemma: Optimizing the Window of Light

Imagine you are an architect tasked with designing a window that is both structurally sound and aesthetically pleasing, but with a twist: you need to maximize the amount of light entering a room. The window is a beautiful composite shape—a rectangle topped with a semi-circular arch.

You have a fixed amount of framing material, which we call the perimeter $P$. This is a classic optimization problem that bridges the gap between pure geometry and the practical application of calculus.

### Phase 1: The Geometry and the Perimeter Trap

Before we dive into the calculus, we must master the geometry. Let the width of the rectangular base be $x$ and its height be $y$.

The semi-circle sits perfectly atop this rectangle, meaning its diameter is $x$, and its radius $r$ is $\frac{x}{2}$.

Now, here is where most students stumble: the perimeter constraint. The problem explicitly states that the perimeter includes the base of the arch. This means the total length of the frame is the sum of the bottom base ($x$), the two vertical sides ($2y$), the curved arc of the semi-circle ($\pi r = \frac{\pi x}{2}$), and the horizontal partition line between the rectangle and the arch ($x$).

Summing these, we get the constraint equation:

$$P = 2y + 2x + \frac{\pi x}{2}$$

This is our foundation. If we miss that partition line, the entire optimization will be skewed.

### Phase 2: The Light Transmission Function

We want to maximize the light $L$. We are told the clear glass in the rectangle transmits three times as much light per square meter as the colored glass in the arch.

Let the transmission rate of the colored glass be $1$ unit. Then the clear glass has a rate of $3$ units.

The total light is:

$$L = 3 \cdot 	ext{Area}_{	ext{rect}} + 1 \cdot 	ext{Area}_{	ext{semi}}$$

Substituting the geometric formulas, where the area of the rectangle is $xy$ and the area of the semi-circle is $\frac{1}{2}\pi r^2 = \frac{\pi x^2}{8}$, we get:

$$L = 3xy + \frac{\pi x^2}{8}$$

To optimize this, we need $L$ in terms of a single variable. From our perimeter equation, we can isolate $y$:

$$2y = P - 2x - \frac{\pi x}{2} \implies y = \frac{P}{2} - x - \frac{\pi x}{4}$$

Substituting this into our light equation transforms it into a function of $x$ alone:

$$L(x) = 3x \left( \frac{P}{2} - x - \frac{\pi x}{4} ight) + \frac{\pi x^2}{8}$$

Expanding this, we find:

$$L(x) = \frac{3Px}{2} - 3x^2 - \frac{3\pi x^2}{4} + \frac{\pi x^2}{8} = \frac{3Px}{2} - 3x^2 - \frac{5\pi x^2}{8}$$

### Phase 3: The Calculus of Optimization

Now, we reach the heart of the problem. To find the maximum, we calculate the derivative $\frac{dL}{dx}$ and set it to zero.

Differentiating our function $L(x)$ gives:

$$\frac{dL}{dx} = \frac{3P}{2} - 6x - \frac{5\pi x}{4}$$

Setting this to zero, we isolate $x$:

$$\frac{3P}{2} = x \left( 6 + \frac{5\pi}{4} ight) = x \left( \frac{24 + 5\pi}{4} ight)$$

Solving for $x$, we get the optimal width:

$$x = \frac{6P}{24 + 5\pi}$$

This is the width that perfectly balances the area of the rectangle and the semi-circle to maximize light transmission.

### Phase 4: The Elegant Conclusion

Finally, we need the ratio of the sides of the rectangle, $\frac{y}{x}$. Substituting our optimal $x$ back into the expression for $y$, we find:

$$y = \frac{P(6 + \pi)}{24 + 5\pi}$$

When we calculate the ratio $\frac{y}{x}$, the complex terms involving $P$ and the denominator $24 + 5\pi$ cancel out beautifully, leaving us with:

$$\frac{y}{x} = \frac{6 + \pi}{6}$$

This result is not just a number; it is the mathematical signature of the perfect window. The ratio of the height to the width of the rectangle must be $6 + \pi : 6$ to ensure the maximum amount of light floods into the room.

It is a testament to how calculus can turn a complex architectural challenge into a simple, elegant ratio.

Visualized Solution

Visualizing the Window Geometry

Understanding the Perimeter Constraint

Expressing $y$ in Terms of $x$

Formulating the Light Transmission Function

Substituting the Area Formulas

Creating a Single Variable Function $L (x)$

The Maximization Condition

Differentiating and Solving for $x$

Finding the Corresponding Height $y$

Calculating the Ratio of the Sides

The Sigma Insight: Maxima and Minima

The Architect's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

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