New Calculus Optimization problem?

Question

A piece of wire 16 inches long is cut into two pieces. One piece is bent to form a square and the other is bent to form a circle. Where should the cut be made in order that the sum of the areas of the square and circle is a minimum? A maximum? (Allow the possibility of no cut)

Answer 1

I don't think that's calculus. Also In the area formula of a circle Pi and the circle are perfect. So in essence the perfect circle can't exist in nature. it's just people obbsessed with perfection. But the cut to maximize it would be made at the diameter of the wire.

Answer 2

We cut on place a, so we have a inch for the square perimeter, which is a/4 for one side of the square and thus a²/16 for the surface of the square. The rest is 16-a inch for the circumference of the circle, which equals 2 pi r, so r = (16-a)/2pi. The surface of a circle is pi r² or for this case (16-a)²/4pi. The sum of the two areas is thus
a²/16 + (16-a)²/4pi. We have to calculate the derivitive to a, which is
a/8 + (2a-32)/4pi and put it zero to find an extremum
=>pi a/2 + (2a-32) = 0 =>a(2+pi/2) = 32 => a=32/(2+pi/2)
or a = 8,96158... and this is a minimum sicede the second derivitive is positive.

Answer 3

1. Let "x" be the length of the piece of wire that is used to make the square. (That means "16–x" will be the length that is used to make the circle.)
2. Express the total area as a function of x:

A = (something with x in it)

3. Take the derivitive of "A" with respect to x.

4. I assume you know how taking a derivitive will help you find maxima and minima. If not, it's time to get off the Internet and hit those books!

Answer 4

You are given a wire of length 16 inches and you are given to cut the wire. The length cut is not specified because we have to find that out.

Suppose the square made from the wire has a perimeter of:
16-x=P

Now suppose the circle has a circumference of x:
x=C

Now we have to use the perimeter and circumference to relate it to their area.

Let's start with the square.
16-x=P
(16-x)/4=S, S=side
((16-x)/4)^2=A A=area

Now lets do the circle
x=2pir
x/2=pir
x/2pi=r
(x/2pi)^2=r^2
pi(x/2pi)^2=pir^2
pi(x^2/4pi^2)=pir^2
x^2/4pi=pir^2
pir^2=A, A=area
x^2/4pi=A

Now you woud add the two areas.
((16-x)/4)^2+x^2/4pi=TotalA, TotalA=total area
(16^2-32x+x^2)/16+x^2/4pi=TotalA
16^2/16-32x/16+x^2/16+x^2/4pi=TotalA
16-2x+x^2/16+x^2/4pi=TotalA

Find the derivative of all of them and set it to 0.
d(16)/dx=0
d(2x)/dx=2
x^2/16=x^2*1/16, Use product rule.
d(x^2/16)/dx=x/8
x^2/4pi=x^2*1/4pi
d(x^2/4pi)/dx=x/2pi

Follow the same exact signs they have and set them to zero, continue the operations.
-2+x/8+x/2pi=0
x/8+x/2pi=-2
Multiply 16pi on both sides.
2xpi+8x=-32pi
pi~3
6x+8x~-96
14x~96
x~6.857142.

*NOTE: The answer is not garanteed to be correct.
It is also an approximation.
x=circumference of circle.

Answer 5

x the length of square
y the length of circle

x+y=16

S(square)=(x/4)^2
S(circle)=pi*(y/(2pi))^2=y^2/(4pi)

f(x,y)=S(square)+S(circle)=(x/4)^2+y^2/(4pi)

f(x)=x^2/16+(16-x)^2/(4pi)

f'(x)=x/8-(16-x)/(2pi)=0
pi*x =64-4x
x=8.96

f(x)=8.96 min

Maximum when all forms circle f(x)=20.32