Question

A wire of length L meters is to be divided into two parts; one part will be bent into a square and the other into a circle. How should the wire be divided to make the sum of the areas of the square and circle as small as possible?

Answer 1

i hate this problem but you are persistent so we can do it i think

Answer 2

lol hey I figured out the maximizing part...

Answer 3

put \(x\) as the amount for the circle, then \(L-x\) is the amount for the square.
circle will have area ... you did?? then your are done

Answer 4

no this is minimizing, the maximizing part was a trick question.
It was zero for the square and L for the circle

Answer 5

Infinitely small circle, rest is square

Answer 6

oh i doubt that

Answer 7

but how do I write that

Answer 8

oooh minimize the area???

Answer 9

Since the area of a circle is always larger than a square for the same perimeter, you would want to minimize the circle

Answer 10

sorry i read it incorrectly
@seattle_ice i think has it, let me shut up

Answer 11

what did you get for the total area in terms of \(x\) ?

Answer 12

(9x-L)/8

Answer 13

wait
that was the derivative, hold on

Answer 14

(x^2)/16 + pi* [ (L-x)/2pi ]^2

Answer 15

hmm i got something different
area of the square is \((\frac{L-x}{4})^2\) and area of the circle is \(\frac{x^2}{4\pi}\) i think

Answer 16

oh i see i switched from what you had

Answer 17

are you talking about the total area or just the circle?

Answer 18

no matter it should work out the same either way

Answer 19

i used \(L-x\) for the perimeter of the square, and \(x\) for the circumference of the circle

Answer 20

Oh, yeah I did the opposite

Answer 21

thought it might be easier having a simpler expression with the \(\pi\) and all

Answer 22

no matter it should work out the same

Answer 23

alright, but I'm still not sure where I went wrong

Answer 24

you get
\[A(x)=\frac{x^2}{16}+\frac{(L-x)^2}{4\pi}\] in your method

Answer 25

derivative would be
\[A'(x)=\frac{4x+\pi x-4L}{8\pi}\]

Answer 26

if you want the critical point, i guess it would be better to write
\[A'(x)=\frac{(4+\pi) x-4L}{8\pi}\] then set
\[(4+\pi)x-4L=0\] and solve for \(x\) in two easy steps

Answer 27

wow, I guess my algebra was wrong

Answer 28

wow, thanks so much! :P really, it would have taken me w while to find my error!

Answer 29

it would be a lot easier to use \(x\) as the circumference of the circle and \(L-x\) as the perimeter of the square

Answer 30

that way you get
\[A(x)=(\frac{L-x}{4})^2+\frac{x^2}{4\pi}\] and taking the derivative is easier

Answer 31

as for the max, you have only one critical point, max occurs there, and min will occur at one of the endpoints of the interval, namely \(x=0\) or \(x=L\)

Answer 32

right, I'll try to rmemeber that for next time.
I realize that setting up the problem can be more importatn than actually doing the problem..

Answer 33

yeah good luck
pretty sure min is make all a square, max is some point in the middle