## A Piece of Wire

#

#### A piece of wire

# You have a piece of plastic-covered clothesline wire one meter long. Your task is to cut the wire into two pieces, one of which you form into a square and the other into a circle, so that the sum of the areas of the two figures is (a) as big as possible; (b) as small as possible.

# Sketch of a solution.

# Let the side of the square be s and let the radius of the circle be r.
Then the perimeter of the square is 4s, and the circumference of the circle is 2*π*r.

The sum of the two perimeters is 100 cm:

4s + 2*π*r = 100.

Solving for s,

s = 25 - π*r/2

# The total area of the two figures is s^{2} + π*r^{2}. Substituting in 25 - π*r/2 for s,

total area A = (25 - π*r/2)^{2} + π*r^{2}.

# Do you know what to do next?

# When the area is at a minimum, r = 7.0012394191 cm. What is s?

When the area is at a maximum, r = 15.91549431 cm. What is s? Why?

Be sure to cut the wire and form your circle and square! You will notice something interesting when you get the figures made.

#

# If the wire were 60 cm in length, how would you cut it?

******************************************************************

# One way to solve the problem:

# We use X for the radius r of the circle. In Y= we enter:

\Y_{1}=(25-πX/2)^{2}+πX^{2}

We want to find where the rate of change of this function is zero.

In Solver,

eqn:0=nDeriv(Y_{1},X,X)

In X= enter some reasonable guess, and press ALPHA SOLVE.

You will see:

X=7.0012394196

From the home screen, enter Y_{1} and you will see

350.06197009

So the total area, when it is at a minimum, is 350 sq. cm.

# Another way:

# Again we use X for the radius r of the circle. In Y= we enter:

\Y_{1}=(25-πX/2)^{2}+πX^{2}

and we set WINDOW

Xmin=0

Xmax=15

Xscl=5

Ymin=300

Ymax=600

We then graph, and use CALC minimum.

# A third way, with lists.

# From the home screen,

# "(25-π*L*X/2)^{2}+π*L*X^{2}"→A

"ΔList(*L*A)/ΔList(*L*X)"→R

# Choose some values for X (radius of the circle in centimeters. For example,

# {5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10}→X

# Now go to STAT Edit and enter X, A, and R.

Look at where the value for A has a rate of change of zero:

# It is between 6.5 and 7.5.

Try again with new values of X between these two values.

# A fourth way (for those who have studied calculus before)

# We have

y=(25-πx/2)^{2}+πx^{2}

Multiply this out and take the derivative by hand:

y=25^{2}-50πx/2+π^{2}x^{2}/4+πx^{2}

=25^{2}-25πx+(π^{2}/4+π)x^{2}

This is a polynomial in x, so

y'=-25π+2(π^{2}/4+π)x

Set this equal to zero, and

25π = 2(π^{2}/4+π)x

Dividing by π, we have

25 = (π/2+2)x

x = 25/(π/2+2)=7.001239419 cm

# A fifth way.

# Let X be the length of the 100 cm that makes the circle, and 100-X be the length that makes the square. Then

circumference c of circle=2πr=x,

so r=x/(2π).

perimeter p of square = 4s = 100-x,

so s = (100-x)/4.

area of circle = A_{circle} = πr^{2} = π(x/(2π))^{2} = x^{2}/(4π)

area of square = A_{square} = ((100-x)/4)^{2}

We want A_{circle} + A_{square} = x^{2}/(4π)+((100-x)/4)^{2} to be as small as possible.

# We enter

\Y_{1}=X^{2}/(4π)+((100-X)/4)^{2}

Then set window:

# and graph:

# Use CALC to find the minimum:

# Or use SOLVER and nDeriv:

**Extra credit.** Cut a 1-meter long wire into two pieces to make a circle and an equilateral triangle such that the sum of the areas of the two figures is as small as possible. Show how you did it.

**More extra credit.** Can you find a general rule for making any regular polygon and circle from a given length of wire, such that the sum of the areas of the two figures is as small as possible?

Webpage Maintained by Owen Ramsey

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