Perimeter Equation: Let's denote the width of the rectangle part of the window as $x$ feet and the height as $y$ feet. The semicircle on top of the rectangle has a radius of $\frac{x}{2}$ feet. The perimeter $P$ of the entire window is the sum of the perimeter of the rectangle $(2x + 2y)$ and the circumference of the semicircle $(\pi \cdot (\frac{x}{2}))$ , but since it's a semicircle, we take half of the circumference. The given perimeter is $30$ feet. $\newline$ So, we have the equation for the perimeter: $\newline$ $P = 2x + 2y + \left(\frac{\pi \cdot (x/2)}{2}\right)$ $\newline$ $30 = 2x + 2y + \frac{\pi x}{2}$ $\newline$ Now, we solve for $y$ in terms of $x$ : $\newline$ $2y = 30 - 2x - \frac{\pi x}{2}$ $\newline$ $y = \left(30 - 2x - \frac{\pi x}{2}\right) / 2$ $\newline$ $y = 15 - x - \frac{\pi x}{4}$
Area Function: Next, we need to find a function that models the area $A$ of the window. The area of the window is the sum of the area of the rectangle ( $x \times y$ ) and the area of the semicircle ( $\pi \times (x/2)^2 / 2$ ). $\newline$ $A(x) = x \times y + (\pi \times (x/2)^2) / 2$ $\newline$ Substitute the expression we found for $y$ : $\newline$ $A(x) = x \times (15 - x - (\pi x)/4) + (\pi \times (x/2)^2) / 2$
Simplified Area: Now, let's simplify the area function $A(x)$ :
$A(x) = x \cdot (15 - x - (\pi x)/4) + (\pi \cdot x^2) / 8$
$A(x) = 15x - x^2 - (\pi x^2)/4 + (\pi x^2) / 8$
$A(x) = 15x - x^2 - (\pi x^2)/4 + (\pi x^2) / 8$
$A(x) = 15x - x^2 - (\pi x^2)/8$
Derivative Calculation: To find the dimensions of the window that admit the greatest amount of light, we need to maximize the area function $A(x)$ . This is done by finding the derivative of $A(x)$ and setting it to zero to find the critical points. $\newline$ $A'(x) = \frac{d}{dx} [15x - x^2 - (\pi x^2)/8]$ $\newline$ $A'(x) = 15 - 2x - (\pi x)/4$ $\newline$ Set $A'(x)$ to zero and solve for x: $\newline$ $0 = 15 - 2x - (\pi x)/4$
Critical Point Calculation: Multiplying through by $4$ to clear the fraction: $\newline$ $0 = 60 - 8x - \pi x$ $\newline$ Combine like terms: $\newline$ $0 = 60 - (8 + \pi)x$ $\newline$ Solve for x: $\newline$ $x = \frac{60}{8 + \pi}$
Maximum Area Verification: Now, we need to check if this value of $x$ gives us a maximum area. We can do this by checking the second derivative or by analyzing the behavior of the first derivative around the critical point. Since this is a typical optimization problem and the area function is a quadratic function that opens downwards (as the coefficient of $x^2$ is negative), we can be confident that this critical point will give us a maximum area. $\newline$ Let's calculate the value of $x$ : $\newline$ $x \approx \frac{60}{8 + \pi}$ $\newline$ $x \approx \frac{60}{8 + 3.14159}$ $\newline$ $x \approx \frac{60}{11.14159}$ $\newline$ $x \approx 5.4$ feet (rounded to one decimal place)
Final Dimensions Calculation: Now that we have the value of $x$ , we can find the value of $y$ using the relationship we derived earlier: $y = 15 - x - (\pi x)/4$ $y = 15 - 5.4 - (\pi \times 5.4)/4$ $y \approx 15 - 5.4 - (3.14159 \times 5.4)/4$ $y \approx 15 - 5.4 - 4.2$ $y \approx 5.4$ feet (rounded to one decimal place)