Question

Calculus optimization problem :
A Norman Window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. find the dimensions of a Norman window of a maximum area if the total perimeter is 28 feet.
x= ? Ft
y= ? Ft

Answer 1

Hey Jackie :)
Ooo this is an interesting problem.

Answer 2

We want to `maximize` area.
So let's see if we can set up a function for our area.

Answer 3

The area is going to be the `area of the rectangular portion` `plus` `the area of the semicircle`.

Answer 4

\[\Large\rm A=xy+\]So there is the rectangular portion.
Do you understand how we'll figure out the area of the semicircle?
What is the radius?

Answer 5

\[\pi/2\]? i think

Answer 6

Hmm they labeled the radius for us.
See the x/2?
The radius ends up being half the length of the rectangle.

Answer 7

Do you remember how to find the area of a circle? :o

Answer 8

for the circle its \[\pi r^2\]

Answer 9

Ok good, and since we have a semi circle, we'll only want half of that \(\Large\rm \dfrac{1}{2}\pi r^2\)

Answer 10

We'll plug our radius in,\[\Large\rm \frac{1}{2}\pi \left(\frac{x}{2}\right)^2\]And simplify,\[\Large\rm \frac{1}{8}\pi x^2\]And add this to rectangle to get the total area of the region.

Answer 11

\[\Large\rm A=xy+\frac{1}{8}\pi x^2\]

Answer 12

We have something to deal with before we can try to `maximize` our area function.

See how it includes x's and y's?
That's no bueno.

We need to use our `constraint` to replace the y with something involving x.

Answer 13

Perimeter = 28.
Let's see if we can come up with a formula for perimeter.

Answer 14

P=2L+2W for the rectangle?

Answer 15

Close. Keep in mind that we don't want to include the top of the rectangle. That has been replaced by the circular portion.

Answer 16

P=2y+x

Answer 17

Mmm k good good good.
And how do we find the distance around a circle? What do we call that? :O
It's a special word.

Answer 18

circumference

Answer 19

Good.
So we'll want to add on `half of the circumference`.

Answer 20

ok so \[c=(1/2)(2\pi r)\]

Answer 21

Yup, so let's plug our radius in again, and simplify.
So it looks like we want to add this \(\Large\rm \frac{1}{2}\pi x\) to our perimeter function.

Answer 22

our perimeter function so far is P=2y + x
now it would be p=2y +x +(1/2) pi x?

Answer 23

Ok good :)
\[\Large\rm P=2y+x+\frac{1}{2}\pi x\]So this is where the constraint comes in handy.
Our perimeter is constrained to 28 feet.\[\Large\rm 28=2y+x+\frac{1}{2}\pi x\]From here we want to `solve for y`.

Answer 24

\[(28-x-(1/2) \pi x)\div 2 =y\]

Answer 25

Ok good.
Let's go ahead and divide each term by 2,
\[\Large\rm \color{orangered}{y=14-\frac{1}{2}x-\frac{1}{4}\pi x}\]

Do you see how we're going to use this information with our area function?\[\Large\rm A=x\color{orangered}{y}+\frac{1}{8}\pi x^2\]

Answer 26

we plug y in and solve for x?

Answer 27

We plug y in, yes.
From there we have successfully established our area function in terms of ONE variable.

Answer 28

\[\Large\rm A=x\left(\color{orangered}{14-\frac{1}{2}x-\frac{1}{4}\pi x}\right)+\frac{1}{8}\pi x^2\]

Answer 29

Now that our Area function is in terms of ONE variable,
we can take it's derivative without any problem.

And from there we can look for critical points, which will lead to max/min points. We're trying to `maximize` our area function, so we're looking for an x value that is a maximum.

Answer 30

Before taking a derivative though....
Uhhh, we better simplify this big mess +_+

Answer 31

x=0 &
\[112/(4+\pi)\]

Answer 32

Hmm I'm not sure where the zero is coming from :d
And my other point came out just a little different than yours.
Let's seeee....

Answer 33

Simplifying, combining like-terms,\[\Large\rm A=14x-\frac{1}{2}x^2-\frac{1}{4}\pi x^2+\frac{1}{8}\pi x^2\]Ehh let's just differentiate from that point,\[\Large\rm A=14-x-\frac{1}{2}\pi x+\frac{1}{4}\pi x\]

Answer 34

\[\Large\rm A'=14-x-\frac{1}{2}\pi x+\frac{1}{4}\pi x\]Woops that should be A' :)

Answer 35

Setting our derivative equal to zero, looking for critical points,\[\Large\rm 0=14-x-\frac{1}{2}\pi x+\frac{1}{4}\pi x\]Multiply both sides by 4,\[\Large\rm 0=56-4x-2\pi x+\pi x\]Combining like-terms,\[\Large\rm 0=56-(4+\pi)x\]

Answer 36

It looks like your second solution was really close, unless I made a boo boo somewhere.

Answer 37

\[\Large\rm x=\frac{56}{4+\pi}\]Something like that, yah? :o

Answer 38

yes i got the same

Answer 39

Ok cool.
Since we only came out with `one` critical point, we can be pretty sure that this is a maximum.

But it doesn't hurt to check I guess.

Answer 40

No no nevermind, let's not check with the number line and all that fancy business... Plugging in that x value would be really annoying :P

Answer 41

so to find y i would plug the value of x into the original equation?

Answer 42

So we've found the length of x that will maximize the area.
Let's plug that x into our `constraint` equation. That's how we'll find our y.

Answer 43

\[\Large\rm y=14-\frac{1}{2}x-\frac{1}{4}\pi x\]

Answer 44

\[y=14-(1/2)(56/4+\pi) -(1/4)\pi (56/4+\pi)\]

Answer 45

Yah.
It looks like it will simplify down quite a bit.
Give it a shot! c:

Answer 46

\[y= 7-4\pi -(\pi^2/4)\]

Answer 47

Hmm 0_o

Answer 48

\[\Large\rm y=14-\frac{1}{2}(x)-\frac{1}{4}\pi(x)\]
\[\Large\rm y=14-\frac{1}{2}\left(\frac{56}{4+\pi}\right)-\frac{1}{4}\pi\left(\frac{56}{4+\pi}\right)\]

Answer 49

Deal with the 1/2 and 1/4 first,\[\Large\rm y=14-\frac{28}{4+\pi}-\pi\left(\frac{14}{4+\pi}\right)\]