# Thread: Minimize Area from cut wire.....

1. ## 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.

2. 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

3. 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.

4. 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

5. 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

6. 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

7. 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: . $\frac{208}{\pi+4}\text{ cm}$

8. ## 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)

9. 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)
.