Q. 16 - Precalculus Enhanced with Graphing Utilities

Question

Norman Windows A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. See the figure. If the perimeter of the window is $20$ feet, what dimensions will admit the most light (maximize the area)?

[Hint: Circumference of a circle $=2\mathrm{\pi r};$ area of a circle $\frac{{\mathrm{\pi r}}^2}{2},$ where $r$ is the radius of the circle.] 

Step-by-Step Solution

Answer

The area of the window is maximized at $L=0.8, w=5.6 ft.$

Step 1. Introduction

First, we need to determine the equation for the perimeter of a semicircle and the rectangle and then we need to determine the dimension that will admit the most light.

Step 2. Solving

The figure, $r$ represents the radius, $L$represents the length, and $w$ represents the width.

The perimeter of the whole window.

$P=w+2L+2\raisebox{1ex}{$\mathrm{\pi r}$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

    $=w+2L+\mathrm{\pi r}.$

      $=2r+2L+\mathrm{\pi r}. \left(\mathrm{Since} \mathrm{w}=2\mathrm{r}\right)\dots \left(1\right).$

The area of the window is given by the equation, $A=Lw+\frac{{\mathrm{\pi r}}^2}{2}.$

   $A=\left(L2r\right)+\frac{{\mathrm{\pi r}}^2}{2}\dots \left(2\right).$

Put $P=20$ in equation (1).

$20=2r+2L+\mathrm{\pi }r.$

$2L=20-2r-\mathrm{\pi }r.$

$L=10-r-\frac{\mathrm{\pi r}}{2}\dots \left(3\right).$

Putting the value of $L$ in equation (2).

$A=\left(10-r-\frac{\mathrm{\pi r}}{2}\right)\left(2r\right)+\frac{{\mathrm{\pi r}}^2}{2}.$

   $=20r-2r^2-\frac{{\mathrm{\pi r}}^2}{2}.$

   $=20r-\left(2+\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)r^2.$

    $=-3.57r^2+20r.$

The $r$ coordinate of the vertex of an equation in the form $a{x}^2+bx+c.$

$r=\raisebox{1ex}{$-b$}\!\left/ \!\raisebox{-1ex}{$2a$}\right..$

  $=\frac{-20}{2·\left(3.57\right)}.$

   $=2.8.$

Substitute $r$ inequation (3).

$L=10-r-\frac{\mathrm{\pi r}}{2}.$

$=10-2.8-\frac{\mathrm{\pi }\left(2.8\right)}{2}.$

$\approx 2.8 ft.$

We know, $w=2r.$

                      $=2\left(2.8\right).$

                       $=5.6 ft.$