Calc Problem

[drummer2233]

A Normal window has the shape of a rectangle surrounded by a semicircle. If the perimeter of the window is 30 ft. express the area A of the window as a function of the width of the window.

- I believe I would start with by using the Area of a Circle and a Rectangle?

 

[Nyithra]

It's a semi-circle, so you would use the area of a circle divided in half.

Adding together the area of the semi-circle and the rectangle will give you the area of the window.

Hope that helps you get started.

 

[tkhunny]

The description is a bit odd. Is your semi-circle sitting on top of the rectangular window or is the rectangular window inside the semi-circle?

 

[drummer2233]

The semicircle is part of the window. Its all one big window, the semicircle is the top part of the window.

 

[tkhunny]

That still didn't clear it up. If the semi-circle sits on the rectangular portion, you should observe that the width of the rectangular section is the same as the diameter of the semi-circle.

 

[soroban]

Hello, drummer2233!

A few malaprops?

A Norman window has the shape of a rectangle surmounted by a semicircle. .If the perimeter
of the window is 30 ft, express the area \(\displaystyle A\) of the window as a function of the width of the window.

I would start with a sketch!

              * * *
          *           *
        *               *
       *                 *

      * - - - - * - - - - *
      |    r         r    |
      |                   |
      |                   |
      |                   | h
      |                   |
      |                   |
      |                   |
      *-------------------*
      : - - - 2r  - - - - :

The perimeter of a semicircle of radius \(\displaystyle r\) is: \(\displaystyle \frac{1}{2}(2\pi r) \:=\:\pi r\)

The perimeter of the window is: .\(\displaystyle \pi r + 2r + 2h \:=\\pi+2)r + 2h\)

The perimeter is 30 ft: .\(\displaystyle (\pi+2)r +2h \:=\:30 \quad\Rightarrow\quad h \:=\:15 - \left(\frac{\pi +2}{2}\right)r\) .[1]


The area of the semicircle is: .\(\displaystyle \frac{1}{2}\pi r^2\)

The area of the rectangle is: .\(\displaystyle (2r)(h) \:=\:2rh\)

The total area of the window is: .\(\displaystyle A \;=\;\frac{1}{2}\pi r^2 + 2rh\) .[2]


Substitute [1] into [2]: .\(\displaystyle A \;=\;\frac{1}{2}\pi r^2 + 2r\left[15 - \left(\frac{\pi+2}{2}\right)r\right] \;=\;\frac{1}{2}\pi r^2 + 30r - \left(\frac{\pi +2}{2}\right)r^2 \)

. . \(\displaystyle A \;=\;\frac{1}{2}\pi r^2 + 30r -\frac{1}{2}\pi r^2 - r^2 \quad\Rightarrow\quad A \;=\;30r - r^2 \) .[3]


Let \(\displaystyle w\) = width of the window.
.. Then: .\(\displaystyle w \,=\,2r \quad\Rightarrow\quad r \,=\,\dfrac{w}{2}\) .[4]

Substitute [4] into [3]: .\(\displaystyle A \;=\;30\left(\frac{w}{2}\right) - \left(\frac{w}{2}\right)^2 \;=\; 15w - \frac{w^2}{4}\)

Therefore: .\(\displaystyle A \;=\;\dfrac{60w - w^2}{4}\)