SOLUTION: A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1116692: A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the perimeter of the window is 32 ft, find the value of x so that the greatest possible amount of light is admitted.
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
2pi%2A%28x%2F2%29%281%2F2%29, to account for the round part of the perimeter.
2pi%2Ax%281%2F4%29
%28pi%2F2%29x

If the other rectangular dimension is y, then accounting for perimeter 32,
%28pi%2F2%29x%2Bx%2B2y=32

Solving that for y:
2y=32-x-%28pi%2F2%29x
y=16-x%2F2-%28pi%2F4%29x


Area A=xy

A=x%2816-x%2F2-%28pi%2F4%29x%29
A=16x-%281%2F2%29x%5E2-%28pi%2F4%29x%5E2
A=16x-%281%2F2%2Bpi%2F4%29x%5E2

.
to continue, finding and setting derivative to 0 would be the best way...
.
.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I can't see your figure; so I guessed that x is the width of the rectangle (and therefore the diameter of the semicircle). And I used y for the height of the rectangle.

We want to find the value of x that makes the greatest possible amount of light admitted through the window, given that the perimeter of the window is 32 feet. That means we want to maximize the area of the window.

So we need an expression in terms of a single variable (preferably x) for the area of the window.

The perimeter of the window, 32 feet, is the width of the rectangle, plus twice the height of the rectangle, plus the circumference of the semicircle:

x%2B2y%2B%28pi%29%28x%2F2%29+=+32

The area of the window is xy%2B%281%2F2%29%28pi%29%28x%2F2%29%5E2+=+xy%2B%28pi%29x%5E2%2F8

We can solve the equation for the perimeter for y in terms of x and substitute into the formula for the area to get the area in terms of x only.

2y+=+32-%281%2Bpi%2F2%29x+=+32-%28%282%2Bpi%29%2F2%29x
y+=+16-%28%282%2Bpi%29%2F4%29x

Then the area of the window in terms of x only is

x%2816-%28%282%2Bpi%29%2F4%29x%29%2B%28pi%29x%5E2%2F8
16x-%28%282%2Bpi%29%2F4%29x%5E2%2B%28pi%29x%5E2%2F8

We differentiate and set the derivative equal to zero to find the value of x that maximizes the area.

16+-+2x%28%282%2Bpi%29%2F4%29%2B%28pi%29x%2F4+=+0
16+-+x+-+%28pi%29x%2F2+%2B+%28pi%29x%2F4+=+0
16+=+x%281%2B%28pi%29%2F4%29
x+=+16%2F%281%2B%28pi%29%2F4%29 = 8.96 ft, to 2 decimal places

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = width and diameter;

    y = height of the rectangle part.


Then the perimeter   

P = x+%2B+2y+%2B+%28pi%2Ax%29%2F2%29  ====>  x+%2B+%28pi%2Ax%29%2F2 + 2y = 32  ====>  y = 16+-+x%2F2+-+%28pi%2Ax%29%2F4.


The area A = xy + %281%2F2%29%2Api%2A%28x%2F2%29%5E2 = x%2A%2816-x%2F2+-+%28pi%2Ax%29%2F4%29 + %28pi%2F2%29%2A%28x%2F2%29%5E2 = 16x - x%5E2%2F2 - %28pi%2F4%29%2Ax%5E2 + %28pi%2F8%29%2Ax%5E2 = -x%5E2%2F2 + 16x - %28pi%2F8%29%2Ax%5E2



Then  the condition for the maximum area  %28dA%29%2F%28dx%29 = 0  takes the form


-x+%2B+16+-+%28pi%2F4%29%2Ax = 0,   or   x%2A%281%2Bpi%2F4%29 = 16  ====> x = 16%2F%281%2Bpi%2F4%29 = 16%2F%281+%2B+%283.14%2F4%29%29 = 8.96 ft.


Answer.  The maximum area is at x = 8.96 ft.

Solved.

==============

Be aware:   The method on how  @josgaritmetic  is doing it   I S   I N C O R R E C T.