##### Find the dimensions of a Norman window of perimeter 20 ft

 Calculus Tutor: None Selected Time limit: 1 Day

Jul 17th, 2015

The Perimeter is distance all round the shape and it is 20ft

The shape is in fom of a tectangle attached to a semi circle which means the perimeter can be given by

p=2y+2x +x *pi

remember the length of a curve is given by a half ( since its a semicircle)  times 2 times pi times the  radius and our radius is x

Let Pi be 3.14

20=2y+2x  + ( 3.14x)

20=2y + 5.14x

hight y= 20/2 - 5.14x/2

y=10 - 2.57x

Area= semicircle + rectangle

A= 1/2 pi * radius ^2 + y *2x

A=0.5 * 3.14 * x^2 + 2xy

Replace the y with 10-2.57x

A= 0.5 *3.14 * x^2 + 2x (10-2.57x)

I will finish the rest in 5 minutes
Jul 17th, 2015

I am working on the last part and will submit it shortly

Jul 17th, 2015

can you please provide a clear y= and x=

Jul 17th, 2015

Let Pi be 3.14

perimeter of semicircle= 1/2 pi * Diameter

=1/2  * 3.14 * 2x

20=2y+2x  + ( 3.14x)

20=2y + 5.14x

hight y= 20/2 - 5.14x/2

y=10 - 2.57x

Area= semicircle + rectangle

A= 1/2 pi * radius ^2 + y *2x

A=0.5 * 3.14 * x^2 + 2xy

A=2xy +1.57x^2

Replace the y with 10-2.57x

2.801122
A=20x -5.14x^2 + 3.14x^2

Group like terms together

A=20x - 3.57x^2

Find the max area by finding the axis of symmetry; x = -b/(2a)

x=-20/ 2(-3.57)

x= 2.80112 ft

y= 10 - 2.57x

y=10-(2.57 *2.80112)

y= 2.801122 ft

Area= 20x -5.14x^2 + 3.14x^2

A=20(2.80112) -5.14 (2.80112)^2 +3.14 (2.80112)^2

A= 56.0224 - 14.39776 + 7.846273

A= 49.47091 Square ft

Jul 17th, 2015

...
Jul 17th, 2015
...
Jul 17th, 2015
Dec 8th, 2016
check_circle