# Minimize Area from cut wire.....

• Oct 19th 2009, 07:47 PM
clayfenderstrat
Minimize Area from cut wire.....
I have been on this problem forever, and cannot solve it. Does anybody have any ideas?

A wire of length 52 cm is to be cut into two pieces. One of the pieces is to be bent into the form of a circle and the other into the form of a square. How should the wire be cut so that the sum of the enclosed areas is a minimum? What is the perimeter of the square? What is the circumference of the circle?

Thanks in advance for all the help.
• Oct 19th 2009, 07:58 PM
tonio
Quote:

Originally Posted by clayfenderstrat
I have been on this problem forever, and cannot solve it. Does anybody have any ideas?

A wire of length 52 cm is to be cut into two pieces. One of the pieces is to be bent into the form of a circle and the other into the form of a square. How should the wire be cut so that the sum of the enclosed areas is a minimum? What is the perimeter of the square? What is the circumference of the circle?

Thanks in advance for all the help.

Call the extreme points of wire A, B and call the cut point C ==> the line AC, say of length $x$, will be shaped bent into a circle and the line CB, of length $52-x$, into a square.

As the circle's perimeter is $x=2r\pi$, with $r=\frac{x}{2\pi}=$ the circle's radius, the circle's area is $\pi\left(\frac{x}{2\pi}\right)^2$, and the square's area is $(52-x)^2$. so the total area of these two shapes is
$\pi\left(\frac{x}{2\pi}\right)^2 + (52-x)^2$.
Now derivate, equal to zero and etc.

Tonio
• Oct 20th 2009, 03:11 AM
chug1
Um just one thing, shouldn't the area of the square be ((52-a)/4)^2, as each side of the square is one quarter of the wire left over from the circle.
• Oct 20th 2009, 03:27 AM
tonio
Quote:

Originally Posted by chug1
Um just one thing, shouldn't the area of the square be ((52-a)/4)^2, as each side of the square is one quarter of the wire left over from the circle.

Of course you're right! Forgot that "tiny" detail.

Tonio
• Oct 20th 2009, 06:07 AM
clayfenderstrat
When I took the derivative, set it equal to zero, then solved, I ended up with the square having a perimeter of (104/3) and the circle having a circumference of (52/3). However, these answers are incorrect. If someone else could solve this and see if they get the same answers, that would be great. I have worked it about 5 times, and get the same thing every time.

Thanks
• Oct 20th 2009, 06:51 AM
tonio
Quote:

Originally Posted by clayfenderstrat
When I took the derivative, set it equal to zero, then solved, I ended up with the square having a perimeter of (104/3) and the circle having a circumference of (52/3). However, these answers are incorrect. If someone else could solve this and see if they get the same answers, that would be great. I have worked it about 5 times, and get the same thing every time.

Thanks

Show your work and let's see what happened: as $\pi$ appears in the expression for $x$ is odd it doesn't appear above.

Tonio
• Oct 20th 2009, 07:01 AM
Soroban
Hello, clayfenderstrat!

Something is wrong . . . Your answers should have $\pi$ in them.

We have: . $A \;=\;\pi\left(\frac{x}{2\pi}\right)^2 + \left(\frac{52-x}{4}\right)^2 \;=\;\pi\left(\frac{x^2}{4\pi^2}\right) + \frac{2704 - 104x + x^2}{16}$

. . . . . . . . $A \;=\;\frac{1}{4\pi}\,x^2 + \frac{1}{16}(2704 - 104x + x^2)$

Then: . $A' \;=\;\frac{1}{2\pi}\,x + \frac{1}{16}(-104 + 2x) \;=\;0 \quad\Rightarrow\quad \frac{1}{2\pi}\,x - \frac{13}{2} + \frac{1}{8}\,x \;=\;0$

Multiply by $8\pi\!:\;\;4x - 52\pi + \pi x \;=\;0 \quad\Rightarrow\quad \pi x + 4x \;=\;52\pi$

Factor: . $(\pi + 4)x \;=\;52\pi \quad\Rightarrow\quad x \;=\;\frac{52\pi}{\pi + 4}$

The circumference of the circle is: . $\frac{52\pi}{\pi + 4}\text{ cm.}$

The perimeter of the square is: . $\frac{208}{\pi+4}\text{ cm}$

• Oct 20th 2009, 07:05 AM
clayfenderstrat
Work for Minimization Problem
y= pi(x/2pi)^2 + ((52-x)/4)^2
y'= 2pi(x/2pi) - 2 ((52-x)/4)
x= 2((52/4)-(x/4))
x=2(13-(2x/4))
x=26-(x/2)
x=(52-x)/2
2x=52-x
3x=52
x=(52/3)
• Oct 20th 2009, 07:23 AM
tonio
Quote:

Originally Posted by clayfenderstrat
y= pi(x/2pi)^2 + ((52-x)/4)^2
y'= 2pi(x/2pi) - 2 ((52-x)/4)

$\color{red}\mbox{This is wrong: you forgot the factor}\,\,\frac{1}{2\pi}\,\,\mbox{from the parentheses in the 1st summand}$

$\color{red}\mbox{you also forgot to multiply by}\,\,\frac{1}{4}\,\,\mbox {in the second summand}$

$\color{blue}Tonio$

x= 2((52/4)-(x/4))
x=2(13-(2x/4))
x=26-(x/2)
x=(52-x)/2
2x=52-x
3x=52
x=(52/3)

.