Chapter 3.4, Problem 16AYU

To determine

To calculate:

The dimensions of Norman window

Answer to Problem 16AYU

The dimensions of Norman window are:

Length: $\frac{20}{\pi +4}$

Height: $\frac{40}{\pi +4}$

Explanation of Solution

Given:

  $p=20ft$

Formula used:

The perimeter of the window is given by the following expression:

  $p=2x+y+2\pi x$

Where,

Sides of rectangle are $2x$ and $y$

The area of the window is given by the following expression:

  $A=xy+\frac{1}{2}\pi x^2$

Calculation:

The perimeter of the window is calculated as:

  $\begin{array}{l}p=2x+y+2\pi x\\ 20=2x+y+2\pi x\end{array}$

  $y=28-\left(\pi +2\right)x$  ...(1)

Now, the area of the window is calculated as:

  $A=x\left[20-\left(\pi +2\right)x\right]+\frac{1}{2}\pi x^2$

  $A=20x-\left(\pi +2\right)x^2+\frac{1}{2}\pi x^2$  ...(2)

Take a derivative of eq. (2):

  $\begin{array}{l}A=20x-\left(\pi +2\right)x^2+\frac{1}{2}\pi x^2\\ A\text{'}=\frac{d}{dx}20x-\left(\pi +2\right)x^2+\frac{1}{2}\pi x^2\\ A\text{'}=20\frac{d}{dx}\left(x\right)+\left(\frac{1}{2}\pi -\pi -2\right)\frac{d}{dx}\left(x^2\right)\\ A\text{'}=20\left(1\right)+\left(\frac{1}{2}\pi -\pi -2\right)\left(2x\right)\left[\begin{array}{l}\because \frac{d}{dx}\left(x\right)=1,\\ \frac{d}{dx}\left(x^2\right)=2x\end{array}\right]\\ A\text{'}=20+\left(\pi +4\right)\left(x\right)\end{array}$

Put: $A\text{'}=0$ then:

  $\begin{array}{l}A\text{'}=20+\left(\pi +4\right)\left(x\right)\\ 20+\left(\pi +4\right)\left(x\right)=0\\ x=\frac{20}{\pi +4}\end{array}$

Now, the variable value of $y$ is calculated as:

  $\begin{array}{l}y=20-\left(\pi +2\right)\frac{20}{\pi +4}\\ y=\frac{20\pi +80-20\pi -40}{\pi +4}\\ y=\frac{40}{\pi +4}\end{array}$

Therefore, the dimensions of the Norman windows to allow the maximum light are:

Length: $x=\frac{20}{\pi +4}$

Height: $y=\frac{40}{\pi +4}$