Question

OPTIMIZATION: A wire 2 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures?

Where should the wire be cut to maximize the total area?

Answer 1

the length of wire used for:

square = 8/(pi+4) m

circle = 2-8/(pi+4) m

Answer 2

but now i want the MAXIMUM for each

Answer 3

@osprey is out of opensudy

Answer 4

he left

Answer 5

i know ho can help though @mathmate

Answer 6

can you help @mathmate

Answer 7

@Amenah8 are you there?

Answer 8

i think so

Answer 9

@bonnieisflash1.0 is your alias Amenah8 ?

Answer 10

@mathmate is it all right to call you puppet master

Answer 11

and NO

Answer 12

@Amenah8
You can tag me when you're online. Good night!

Answer 13

she's here

Answer 14

@mathmate sorry my computer messed up!

Answer 15

maximum area?

Answer 16

my computer is slow

Answer 17

@Amenah8 no problem.
You want the maximum total area, right?

Answer 18

bye

Answer 19

i have a guess

Answer 20

@mathmate you got a guess?

Answer 21

Nope, tell me about it!
@satellite73

Answer 22

the question was Where should the wire be cut to maximize the total area? so i guess the max for each, the square and the circle

Answer 23

my guess is to maximize, use all for the circle
should get the biggest area that way

Answer 24

but if you did the "min" part you already have the function and the derivative and the min,
use the other critical point if there is one, and also try the endpoints

Answer 25

what do u mean?

Answer 26

you want to maximize and area with a fixed perimeter, make a circle
it gives the biggest area

Answer 27

@satellite73 Agree!
If we split it up more or less half/half, it would be min.

Answer 28

did you get a function for the combined areas?

Answer 29

@satellite73
I do, not sure if @Amenah8 did.

Answer 30

@Amenah8
Have you been following @satellite73 posts?

Answer 31

BTW, @Amenah8 Have you done calculus yet?

Answer 32

if you got some solution you must have a a function that gave the area if some amount was used for the circle]
otherwise you couldn't have done it

Answer 33

@Amenah8
Are you there at the keyboard?

@satellite73 Yeah, the square takes slightly more wire than the circle.

Answer 34

@satellite73
But perhaps OP has not done calculus. The question could be just to illustrate that Area is proportional to linear dimensions squared.

Answer 35

no i'm in calculus now. it's supposed to be optimization --> i have the mins but i still need the max

Answer 36

hold on i can paste my work when i found the min

Answer 37

Good, read through the thread, you'll find the answer from @satellite73 's posts.
If you have questions as to why, tag @satellite73 or me, your choice.

Answer 38

derivative was (1/8)x - (1/2pi)(2-x)

Answer 39

and square min was 8/(pi+4) and circle min was (2-8)/(pi+4)

Answer 40

but how do i use that to find the max for each?

Answer 41

@satellite73

Answer 42

you said to split them in half?

Answer 43

no
check the endpoints of the interval of your function, which i still don't see

Answer 44

i see
"derivative was (1/8)x - (1/2pi)(2-x)" but i don;t see what the original function was

Answer 45

oh hold on

Answer 46

oh here we don't need it

Answer 47

A for circle and square combined = (1/16)x^2 + pi(2-x / 2pi)^2

Answer 48

the critical points are
a) the zero of the derivative of
b) where the derivative is undefined,
i.e where the denominator is 0`

Answer 49

or the endpoints of the interval of the domain namely \(0\) and \(2\)

Answer 50

so plug those into the original equation?

Answer 51

@satellite73
For your info, both expressions for area and derivatives correspond to mine.

Answer 52

ok so good enough , check the endpoints of the interval, which are also where the derivative is undefined
my guess is the one that says "all circle" is the biggest

Answer 53

@satellite73
Right!
http://prnt.sc/czfn0f

Answer 54

ohhhh this makes much more sense!

Answer 55

but it does not show a max

Answer 56

Remember \(global\) max can occur at any of the critical points, or the closed endpoints of the domain.

Answer 57

so about when y = 2.5?

Answer 58

is this your original function
(1/16)x^2 + pi(2-x / 2pi)^2

Answer 59

yes

Answer 60

is it clear what the endpoints of the interval are?

Answer 61

i mean endpoints of the domain

Answer 62

Welcome back @satellite73 :D

Answer 63

thanks !

Answer 64

wasn't the endpoint 0 and 2?

Answer 65

yes

Answer 66

check them

Answer 67

.32 at 0 and .25 at 2

Answer 68

then 3.2 it is

Answer 69

and that is the max for the square or the circle?

Answer 70

seems wrong to me though if you make a circle of radius two the area is \(4\pi\)

Answer 71

it is not the max for a square and a circle
it is the max combined area

Answer 72

so then what is the max for each separately?

Answer 73

@satellite73 sorry my computer messed up

Answer 74

@Amenah8
`.32 at 0 and .25 at 2`
are the 'total' areas.
The first one refers to x=0 (left extreme) where the size of the square is zero.
So the circle gets a \(circumference\) of 2 metres. From there, you would extract the radius, given by 2\(\pi\)r=2, or r=1/\(\pi\) to give an area of \(A_{total}=0+\pi(1/\pi)^2\)=0.32.
Similarly, 0.25 is at x=2, giving a circle of 0 circumference and a square of 2 metres in perimeter. The side is therefore 2/4=0.5, hence \(A_{total}=(1/2)^2+0=0.25\)
It's up to you to choose the global maximum from all of the above, possibly with the help of the graph. Mathematically it should be the maximum among all critical points and end-points of domains (if applicable).
Hope that helps to clear up queries.