Answer to Question #136676 in Calculus for Kare

Question

A norman window with a perimeter of 20 units has the shape of a rectangle surrmounted by a semi circle. If its base is x units long and the height of rectangle is y units long, write a mathematical model for its area in term of x

Answer 1

$\displaystyle\begin{aligned}\textsf{Perimeter}\hspace{0.1cm} &=\hspace{0.1cm} \textsf{Length of the semi-circle}\\&+ 2 \times \textsf{breadth} + \textsf{length} \end{aligned}\\ \begin{aligned} \textsf{Length of the semi-circle} &= \pi r\\ \therefore \pi r + 2y + x = 20 \end{aligned}\\ r = \frac{1}{2}\times \textsf{the length of the rectangle} = \frac{x}{2} \\ \therefore \frac{\pi x}{2} + 2y + x = 20\\ \pi x + 4y + 2x = 40\\ \textsf{Writing}\hspace{0.1cm} y \hspace{0.1cm} \textsf{in terms of}\hspace{0.1cm} x,\\ 4y = 40 - (\pi + 2)x\\ y = 10 - \frac{(\pi + 2)x}{4}\\ \textsf{Area of the norman window} =\\\textsf{Area of the circle} + \\\textsf{Area of the rectangle} \\\therefore A = \frac{\pi r^2}{2} + xy = \frac{\pi x^2}{8} + xy \\ \begin{aligned} \Rightarrow A &= \frac{\pi x^2}{8} + 10x - \frac{\pi + 2}{4}x^2\\ &= \frac{\pi x^2 + 80x - 2(\pi + 2)x^2}{8}\\&\frac{(\pi - 2\pi - 4)x^2 + 80x}{8}\\&\frac{80x - (\pi + 4)x^2}{8} \end{aligned}\\ \therefore \textsf{The area of the norman window}\\\textsf{in terms of}\hspace{0.1cm} x \hspace{0.1cm}\textsf{is}\hspace{0.1cm} \frac{x(80 - (\pi + 4)x)}{8}$