Question

A Norman Window had the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30ft, find the dimensions of the window so that the greatest possible amount of light is admitted. (maximize area)
http://imageshack.us/photo/my-images/163/normanwindow.jpg/

Answer 1

2r = x

Answer 2

yeah I got that...
so A = xy + (1/2)pi (x/2)^2 ?

Answer 3

Now get perimeter equation.

Answer 4

yes

Answer 5

P = 2y + x = (1/2)x pi = 30 ?

Answer 6

* 2y + x + (1/2)x pi

Answer 7

okay, solve for \(y\) and substitute it into the area function. Now all you have to do is find the maximum of a single variable function.

Answer 8

first of all let
x = the width of the window and diameter of the semicircle
h = height of the rectangular portion of the window
2h + x + .5x*pi = 30

Answer 9

mmm
so y = 15 - (1/2)x + (1/4)x pi ?

Answer 10

area of a semi-circle = 1/2* pi * r^2

Answer 11

where'd the 30 go?

Answer 12

Oh please let me know if I'm doing something wrong :P

Answer 13

h = (30-x - 0.5x * pi)/2
Total Area = semicircle + rectangle

Answer 14

ok so now plug in values in area of semi circle formula

Answer 15

A = x(15 - (1/2)x + (1/4)x pi) + (1/2)pi (x/2)^2 ?
this is complicated 0.o

Answer 16

=1/2* pi * r^2 + h * l
= 0.5 * pi * (0.5x)^2 + ((30-x - 0.5x * pi)/2) * x

Answer 17

^ thats total area

Answer 18

anyways @jigglypuff314 you need to simplify that eq

Answer 19

erm so A = 15x - 0.5 x^2 + (5/8) x^2 pi

Answer 20

A = 0.3925x^2 + 15x -2.355x^2
A = -1.9625x^2 + 15x

Answer 21

Don't throw out \(\pi\). Keep it alive.

Answer 22

@jigglypuff314 Do you got this, or are you afraid of algebra?

Answer 23

I might get cross eyed and mess up something somewhere :P

Answer 24

finding the axis of symmetry:
x = -b/(2a)
x = -15 / (2*-1.9625)

Answer 25

would it be A = 15x - 0.5 x^2 + (5/8) x^2 pi
so that A' = -x + (5/4)pi x + 15 ?

Answer 26

oh pi has a value that can be multiplied (3.1415)

Answer 27

x = -15 / -3.925
x = 3.821
rest is put value of x back in the total area formula
A = 0.5 * pi * (0.5x)^2 + ((30-x - 0.5x * pi)/2) * x

Answer 28

I've done this question in 8th class I was 12-13 then

Answer 29

Area will be 44.28 units^2

Answer 30

erm
what happened going from
A = 0.5 * pi * (0.5x)^2 + ((30-x - 0.5x * pi)/2) * x
A = 0.3925x^2 + 15x -2.355x^2

Answer 31

umm I simplified it

Answer 32

pi is not a variable you can multiply it with coefficient

Answer 33

Get in the habit of not multiplying constants in. No one really wants you to do that.

Answer 34

my brain hurt at looking at it sorry xD
so ((30-x - 0.5x * pi)/2) * x = (15 - 0.5x - 0.25xpi)x
= 15x - 0.5x^2 - 0.25pi x^2
= 15x - 0.5x^2 - 0.785398x^2
= 15x - 1.285x^2

Answer 35

annas, calm down :) I'm sure wio didn't mean any harm >,<
I like how you do things :)

Answer 36

Uhh, sorry about that. I was just being stupid.

Answer 37

well you're in charge of helping me now, so help :P

would it be A = 15x - 0.5 x^2 + (5/8) x^2 pi
so that A' = -x + (5/4)pi x + 15 ?

Answer 38

@wio ?

Answer 39

You want me to check work? I'm not good at that.

Answer 40

\[
A = xy + \frac{1}{2}\pi \left(\frac{x}{2}\right)^2
\]\[
P = 30 =2y + x + \frac{x}{2} \pi
\]

Answer 41

\[
y = 15 - \frac {x}{2} - \frac{x}{4}\pi
\]

Answer 42

\[
A = x\left(15-\frac x2 - \frac x4\pi\right) + \frac 12 \pi \left(\frac x2\right)^2 = 15x-\frac {x^2}2 -\frac {x^2}4\pi +\frac 18\pi x^2
\]

Answer 43

\[
A' = 15 - x - \frac{x}{2}\pi -\frac{1}{4}\pi x =0
\]

Answer 44

\[
15 = \left(1+\frac 12\pi +\frac 14\pi\right)x
\]

Answer 45

\[
60 = (4+2\pi +\pi)x = (4+3\pi)x\implies x=\frac{60}{4+3\pi}
\]

Answer 46

ok, I see what I did wrong now :) thanks :)

Answer 47

\[
y = 15 - \frac{30}{4+3\pi} - \frac{15\pi}{4+3\pi} = \frac{15(4+3\pi)}{4+3\pi} - \frac{30+15\pi}{4+3\pi}=\frac{30+30\pi}{4+3\pi}
\]

Answer 48

I may be wrong.

Answer 49

would it be 60 = (-4 + 2pi + pi)x
instead of 60 = (4 + 2pi + pi)x ?

Answer 50

I don't know. I mix things up.

Answer 51

oh well :P
Thanks for your help! :)