Optimization Problem

**BClary** · November 18th 2007, 11:41 AM

Here is the problem,

A straight piece of wire is to be cut into two pieces. One piece is to be formed into a circle and the other piece into a square. Where should the wire be cut so that the sum of the areas of the circle and square will be greatest? The least?

I know the square area formula x=side, x(squared). The same stuff for the circle, x=circumference, x=2(Pi)r

I do not know the Length of the wire, but that length is a constant away (right?, I will assume this for now) so I will use “L”.

So the side of the square will be x, the circumference will be L-x

Stuck right here, not sure what to do next

**Opalg** · November 18th 2007, 11:58 AM

Originally Posted by BClary

A straight piece of wire is to be cut into two pieces. One piece is to be formed into a circle and the other piece into a square. Where should the wire be cut so that the sum of the areas of the circle and square will be greatest? The least?

I know the square area formula x=side, x(squared). The same stuff for the circle, x=circumference, x=2(Pi)r

I do not know the Length of the wire, but that length is a constant away (right?, I will assume this for now) so I will use “L”.

So the side of the square will be x, the circumference will be L-x

Stuck right here, not sure what to do next

Good start, except that the perimeter of the square will be x (not the side). So the side of the square will be x/4, and the radius of the circle will be (L–x)/2π.

That means that the total area enclosed will be $(x/4)^2 + \pi\Bigl({\textstyle\frac{L-x}{2\pi}}\Bigr)^2$ . Now use calculus to find when that is a maximum or a minimum.

**BClary** · November 18th 2007, 12:10 PM

wow, thats for the great start, looks like I am just a "little" algebra away now

**BClary** · November 18th 2007, 06:24 PM

Okay, found the algebra easier if I made the square "L-x" and let the circle be "x"

so

the sum of the two areas would be

Circle + Square = 2Pi(x/2Pi)(squared)+((L-x)/4)(squared)

I think that is right. So the next step should be to take a derivative, SO I will do the left hand side first, add the right hand later (this can get ugly looking quickly here):

Left hand side:
A'(x)=[(Pix(squared))'(4Pi(square))-Pix(squared)(4Pi(squared)]/16Pi(squared)

[(2xPi)(4Pi(squared))-0]/(16Pi(fourth))

8xPi(third)/16Pi(fourth)

x/2Pi for the left hand side

right hand now:

A'(x)=((L-x)/4)(Squared)

2(L-x)(-1)(16)-0/16

-2(L-x)/16

-(L-x)/8

Combined back:

x/2Pi + -L+x/8

(4+xPi-Lpi)/8Pi

so set top to zero

4+xPi-LPi=o

Stuck again, been at this for hours, I know I need to find both max and min, but the L is causing me great problems. The original problem does not address L or ask for it, could I use a simple value of 1 "one" for it, as the ratio of any cut of the wire would apply to all lengths. I need to walk away for a moment, bang my head against a solid object, multiple times (no I will not take a derivative to find v of that), and hope someone reads this a gives me the push in the right direction

Signed
Soon to have a headache

**angel.white** · November 18th 2007, 08:04 PM

Edit: I figured I should add an overview to make this easier to follow since it's so long.
OVERVIEW
1. Set up the problem (define values and default relationships)
2. Solve for area of a square in terms of length of it's wire
3. Solve for area of a circle in terms of length of it's wire
4. Add them together to get a formula for total area
5. Find where the slope of the area = 0
6. Test that point.
7. Test the end points of your domain to see where they lie
8. Wrap up the process into a conclusion

SETUP THE PROBLEM
the length of the wire is L
the length of the piece that will make the square is s
the length of the piece that will make the circle is c
c+s=L
c=L-s

FIND THE AREA OF THE SQUARE
So, now you need to find an equation to make the square, you know the circumference of a square is 4 times one side, and that is equal to the length of your piece of wire which will make the square. So one side of the square will be 1/4 of s, or s/4. And since the Area of a square is the one side squared, you know that the area of the square is $(\frac{s}{4})^{2}$ which is $\frac{s^{2}}{16}$

FIND THE AREA OF THE CIRCLE
Your piece of wire b will become the circumference of the circle, and the circumference of a circle is $2\pi r$ so $c=2\pi r$ and so $r=\frac{c}{2\pi}$ . And you know the area of the circle is $\pi r^{2}$ so substitute the value of r into the equation for area and get $\pi(\frac{c}{2\pi})^{2}$ Which can be simplified to $\pi(\frac{c^{2}}{4\pi^{2}})$ Which simplifies to to $\frac{c^{2}}{4\pi}$

Now, you don't want two variables (s and c) and we know that c=L-s (L is not a variable, it is a constant). so we substitute that into our equation.
$\frac{(L-s)^{2}}{4\pi}$ and simplify: $\frac{L^{2}-2Ls+s^{2}}{4\pi}$

FIND THE EQUATION FOR THE SUM OF THE SQUARE AND THE CIRCLE
This is easy, just add the area of the square to the area of the circle
$\frac{s^{2}}{16}+\frac{L^{2}-2Ls+s^{2}}{4\pi}$

Common denominator:
$\frac{\pi s^{2}}{16\pi}+\frac{4(L^{2}-2Ls+s^{2})}{4(4\pi)}$

Simplify:
$\frac{\pi S^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$

And turn this into an equation, by realizing it is equal to our area, A, in terms of s (s is the only variable in this equation since we substituted out the b and L is a constant)
$A(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$

CONSIDER
A=area in terms of s, so the value of A will give us our total area, and the graph of A will show us our area as it changes according to a. Now, we want the point where area is greatest. At that point, we know that it will be the highest on the graph of area. And we know that because it is highest, it must be higher than all the points around it, which means the slope of A is increasing up to it, and decreasing after it, and equal to 0 at that point.

By the same thinking, wherever the area is minimized, the slope will be decreasing before that piont and increasing after it, and equal to zero on it.

So wherever the slope is equal to zero is a potential maximum/minimum value of area. So lets find the slope and set it to zero, then test the points around it to see what the slope is doing.

So we want to find the equation for the slope of this line, and then set that equal to zero. The derivative is the equation for the slope of the line, so lets find the derivative.

FIND THE DERIVATIVE
Initial equation:
$A\prime(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$

Rewrite:
$A\prime(s)=\frac{1}{16\pi}(\pi s^{2} + 4L^{2}-8Ls+4s^{2})$

Differentiate (remember that L is a constant, and the derivative of a constant is zero)
$A\prime(s)=\frac{1}{16\pi}(2\pi s -8L+8s)$

FIND THE ZERO OF THE SLOPE
Now we have the equation of our slope, lets find where it is equal to zero as that will be a potential maximum value:
$0=\frac{1}{16\pi}(2\pi s -8L+8s)$

Divide out the irrelevant constant
$0=2\pi s -8L+8s$

Divide out a 2
$0=\pi s -4L+4s$

Factor out an s
$0=s(\pi +4) -4L$

Add 4L
$4L=s(\pi +4)$

Divide by the coefficient of s
$\frac{4L}{\pi +4}=s$

CONSIDER
Now we have a value for s where the change in the total area of our circle and square is zero, but lets make sure that it is a maximum. Because all of the terms we are using are arbitrary, (s, c, L) it is very difficult to choose actual points to plug into our first derivative, so what we can instead do is find the second derivative, plug our point into the second derivative and see whether it is concave up or down. If it is concave down, then the open end is down, so it must be increasing up to that point and decreasing after it, making it a maximum. If the open end is up, it looks like a parabola, and we can see that our point is at the bottom, making it a minimum.

FIND THE SECOND DERIVATIVE
First derivative
$A\prime(s)=\frac{1}{16\pi}(2\pi s -8L+8s)$

Differentiate:
$A\prime\prime(s)=\frac{1}{16\pi}(2\pi +8)$

Simplify
$A\prime\prime(s)=\frac{\pi +4}{8\pi}$

CONSIDER:
This tells us that Now we can see that the second derivative is ALWAYS positive, this means that the slope is always concave up, so our s value must be a minimum. This tells us that when $s=\frac{4L}{\pi +4}$ our area is at a minimum. Well what do we do now? what about our maximum? There was only 1 value where the prime was equal to zero, so where does our other point come from?

Well, we only have 2 other critical numbers, the two endpoints of our graph, where the domain begins and ends. Now we know we can't have a length less than zero, so s must be > zero. And we can't have a length of s that is greater than the length of the wire, so s is < to L. This means that the domain is 0 < s < L

Now we have two more critical values to check out. Whichever is higher must be our maximum value.

CHECK THE AREA OF OUR TWO ENDPOINTS
Initial equation
$A(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$

*Check area of s=zero
$A(0)=\frac{\pi 0^{2} + 4L^{2}-8L*0+4*0^{2}}{16\pi}$

Simplify
$A(0)=\frac{4L^{2}}{16\pi}$

Simplify
$A(0)=\frac{L^{2}}{4\pi} \approx .079599L^{2}$

*Check area of s=L
$A(L)=\frac{\pi L^{2} + 4L^{2}-8L*L+4L^{2}}{16\pi}$

Simplify
$A(L)=\frac{\pi L^{2}}{16\pi}$

Simplify
$A(L)=\frac{L^{2}}{16} = .0625L^{2}$

CONSIDER
Now our two endpoints are each in terms of L^2, and we know that L is positive, so we can just compare their coefficients and see which is greater: .079599 > .0625

Therefore our area is maximized when s equals 0

CONCLUSION
We know our area is minimized when $s=\frac{4L}{\pi +4}$
And maximized when $s=0$

So to get the greatest area, do not cut the wire at all and instead use it to make only a circle, and to get the least area, cut the wire $\frac{4}{\pi+4}$ of the length and use that to make a square, and the remaining bit to make a circle.

----
There was a lot in there, I hope I didn't make any errors :/

**BClary** · November 19th 2007, 02:04 AM

If you are not a teacher, you should be, great explaination, its earily, but I will review and review, thank you

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