# Thread: Norman window and greatest possible area of a rectangle

1. ## precalculus problems

Hi everybody I'm a precalculus student, and I'd like some help to understand these two problems?

1 A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=1-x2. What are the dimensions of such a rectangle with the greatest possible area?

2-A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 26 feet?

2. Hello, skorpiox!

2) A Norman window has the shape of a semicircle atop a rectangle
so that the diameter of the semicircle is equal to the width of the rectangle.
What is the area of the largest possible Norman window with a perimeter of 26 feet?
Code:
              * * *
*           *
*               *
*                 *

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

The circumference of the semicircle is: $\pi r$
The perimeter of the rectangle is: $2r + 2h$

The total perimeter is 26 feet.
. $\pi r + 2r + 2h \:=\:26 \quad\Rightarrow\quad h \:=\:13-\frac{\pi+2}{2}r$ .[1]

The area of the semicircle is: $\tfrac{1}{2}\pi r^2$
The area of the rectangle is: $2rh$

The total area is: . $A \;=\;\tfrac{1}{2}\pi r^2 + 2rh$ .[2]

Substitute [1] into [2]: . $A \;=\;\tfrac{1}{2}\pi r^2 + 2r\bigg[13-\frac{\pi+2}{2}r\bigg]$
. . which simplfies to: . $A \;=\;26r - \frac{\pi+4}{2}\,r^2$

And that is the function we must maximize . . .

3. Hello,

1 A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=1-x2. What are the dimensions of such a rectangle with the greatest possible area?
So let a and b the abscissas the base :

Code:
  |
|
|
_|___,______,____
|    a      b
|
The upper corners points C and D are on the parabola.
Consider that C has an abscissa of b and D an abscissa of a (since it's a rectangle, with a base on the x-axis, they have the same abscissa)
b>a

C has coordinates (b,1-b²) and D : (a,1-a²) (remember that C and D are on the parabola)
But since it's a rectangle, CD is parallel to AB, and thus parallel to the x-axis. Hence C and D have the same ordinate
So in fact D has coordinates (a,1-b²)

Hence 1-b²=1-a²
---> a²-b²=0 ---> (a-b)(a+b)=0
So either a=b, which would make an area of 0, either a=-b.
---> a=-b.
Since b is the greatest, b is positive.

The area is given by $AB \times BC$
Distance AB=(b-a)=2b
Distance BC=1-b²

So $A(b)=2b(1-b^2)$
This is the function to maximize.

Remember that b is between 0 and the positive y-intercept of the parabola (1-x²=0 --> x=1) (because the rectangle is inscribed and because b>0)

4. Originally Posted by skorpiox
Hi everybody I'm a precalculus student, and I'd like some help to understand these two problems?

1 A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=1-x2. What are the dimensions of such a rectangle with the greatest possible area?

2-A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 26 feet?

1) $A = 2xy = 2x(1-x^2)$ ... locate the vertex of the parabolic graph representing area to find the maximum.