min. Areas, Given Length of Wire for perimeter

**mike_302** · February 27th 2009, 09:03 PM

A friend and I have tried this question waaaay too long. It's calculus and we cannot solve this dumb question to get the right answer !

Question: A piece of wire 75m long is divided into two pieces. One piece is used to form a circle. The other piece is used to make a square. Find the lengths of the pieces that maximize the total area.

Our work: We have two pieces --> length y and (75-y)

Psquare = (75-y)
side of the square is (75-y)/4

Circum. circle = 2pi(r) , y/2pi=r

Asquare = side^2

Acircle=(y^2)/4

You see I did quite a bit of work... basically, I subbed in to minimize the following eq'n:
pi(r)^2+s^2
= (y^2)/4 + ((75-y)^2) /16

and I keep getting weird numbers.

Pleeease, someone tell me what is wrong here because we have checked over and over for hours!

**mr fantastic** · February 27th 2009, 09:18 PM

Originally Posted by mike_302

A friend and I have tried this question waaaay too long. It's calculus and we cannot solve this dumb question to get the right answer !

Question: A piece of wire 75m long is divided into two pieces. One piece is used to form a circle. The other piece is used to make a square. Find the lengths of the pieces that maximize the total area.

Our work: We have two pieces --> length y and (75-y)

Psquare = (75-y)
side of the square is (75-y)/4

Circum. circle = 2pi(r) , y/2pi=r

Asquare = side^2

Acircle=(y^2)/4

You see I did quite a bit of work... basically, I subbed in to minimize the following eq'n:
pi(r)^2+s^2
= (y^2)/4 + ((75-y)^2) /16

and I keep getting weird numbers.

Pleeease, someone tell me what is wrong here because we have checked over and over for hours!

Read this thread: http://www.mathhelpforum.com/math-he...n-problem.html

All you have to do is change a couple of numbers.

**mike_302** · February 27th 2009, 09:21 PM

Uhhhhh... embarrassing..
We figured it out. But we are so annoyed by this question: WE need to know what the reasoning is for our problem, which we think we have figured out.

We kept getting one certain number before because we were doing the question this one way: We were always working y out to be the radius of the circle (in my last post I said we were using y and (75-y) because on our last attempt, which i screwed up the math, that was my newest and most reasonable approach)

When we work y to be the perimeter of one, and then "related to the perimeter" of the other, it all works out.

But ..... There was no question that when we had y=radius, the way we worked out the question, it was all right... So, I am trying to figure out fundamentally, why using y as the radius, vs. using y as the permieter, and (75-y) as the other perimieter... only one way works.

Sorry

!!

**mike_302** · February 27th 2009, 10:11 PM

MAJOR EDIT:

By some MASSIVE fluke, I put the numers in wrong and managed to obtain the answer that as in the back of the book, so disregard my last message.

So I took a look at the link you gave me, and we followed it for the past hour or so --- line by line, letter by letter, symbol by symbol . My solution is the EXACT same . but it does NOT give the right answers. And the book DOES have the right answers because we subbed in their numbers to find the areas, and they DO produce the lowest area, and the two numbers add to 75.

so this is unsolved so far! I am absolutly in shock as to how I managed to receive the correct answer by fluke, by entering numbers into the calculator wrong.

**mr fantastic** · February 28th 2009, 01:31 AM

Originally Posted by mike_302

MAJOR EDIT:

By some MASSIVE fluke, I put the numers in wrong and managed to obtain the answer that as in the back of the book, so disregard my last message.

So I took a look at the link you gave me, and we followed it for the past hour or so --- line by line, letter by letter, symbol by symbol . My solution is the EXACT same . but it does NOT give the right answers. And the book DOES have the right answers because we subbed in their numbers to find the areas, and they DO produce the lowest area, and the two numbers add to 75.

so this is unsolved so far! I am absolutly in shock as to how I managed to receive the correct answer by fluke, by entering numbers into the calculator wrong.

Originally Posted by Moo and plagiarised by Mr F

Hello,

This is tricky, or I didn't get the thing

Let x be the length of the wire bent to a circle and y to form a square.

Let's study the circle

x is its perimeter, which is also $2 \pi R$ , where R is the radius. According to it, we have the equation $x=2 \pi R \Longleftrightarrow R=\frac{x}{2 \pi}$
The area enclosed by the circle is $A_1=\pi R^2$ , which is also $A_1=\pi \left(\frac{x}{2 \pi} \right)^2 \Longleftrightarrow { \color{red} A_1 = \frac{x^2}{4 \pi} }$

Let's study the square

$y= {\color{blue}75} -x$ is its perimeter.

If q is the length of the side of the square, the perimeter is 4q.
Hence $y=4q \Longleftrightarrow q=\frac{y}{4}$

The area of a square is $A_2=q^2 \Longleftrightarrow A_2=\frac{y^2}{4^2} \Longleftrightarrow {\color{red} A_2=\frac{({\color{blue}75}-x)^2}{16}}$ .

Total area

$A=A_1+A_2=\frac{x^2}{4 \pi}+\frac{({\color{blue}75}-x)^2}{16}=\frac{x^2}{4 \pi}+\frac{{\color{blue}5625 - 150 x} + x^2}{16}$ $= x^2 \left(\frac{1}{4 \pi}+\frac{1}{16} \right)+ \frac{{\color{blue}5625 - 150x}}{16}$

So find the derivative and see, in the interval [0, 75] (since the wire is 75 cm long) where the maximum is and where the minimum is

I did this and got $x = \frac{75 \pi}{\pi + 4}$ .

**mike_302** · February 28th 2009, 07:17 AM

oh boyo.

Okay, it took some sleep to finally understand what went on in that last step (up until then, we had the same math).... The way I was doing at the last step was getting the same denominator of 16[pi] on ALL of the terms, then do my derivative from there.... I guess you cannot do that.

Anyways, my question now would be: How do you take the derivative in this case? we have only gone as far as the product rule, and, if memory serves me right, this form of equation takes some other rule...

So could someone walk me through finding the derivative after the last step?

Extreme thank you to mr fantastic !

**mike_302** · February 28th 2009, 07:42 AM

my GOODNESS I FOUND IT! I was doing EVERYTHING RIGHT AND IT WASN'T A FLUKE!

I wrote DOWN (1/2)[pi] in my very last step before solving for the answer, but when I first entered into the calculator, I entered (1/2[pi]) , as it should be, and I received the correct answer.... But when we tried to duplicate that, and decided it was a fluke, we were reading EXACTLY what was on my page and entering that... Couldn't get the right answer. Problem was: My writing --> I put the [pi] just a littttle too far from the 2 for me to believe it was 2*pi .... so i thought it was 1/2 * pi

My condolences to anyone that has to put up with my stupidity :P

Thanks though! To everyone (for putting up with this moron I call "me")

**yashas** · February 28th 2009, 08:27 AM

it depends on how u want to divide the wire.
as the area must be maximum the square must be of the largest side .....

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