SOLUTION: A piece of wire 60cm in length is cut and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combi

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A piece of wire 60cm in length is cut and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combi      Log On


   

Question 334785: A piece of wire 60cm in length is cut and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combined area of the circle and the square.
I know to solve this you need to make an equation in which the X (the amount of wire for a square) is a function of the Total area, but stuck
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!




Circumference of circle + Perimeter of Square = 60

2%2Api%2Ar+%2B+4x%22%22=%22%2260

Solve for r

2%2Api%2Ar%22%22=%22%2260-4x

Divide through by 2

pi%2Ar%22%22=%22%2230-2x

Multiply both sides by 1%2Fpi

r%22%22=%22%221%2Fpi%2830-2x%29

Let y = Area of circle + Area of square

    y=pi%2Ar%5E2%2Bx%5E2

We need to substitute for r%5E2

r%22%22=%22%221%2Fpi%2830-2x%29
r%5E2%22%22=%22%221%2Fpi%5E2%2830-2x%29%5E2

Substituting:

y=pi%2A%22%221%2Fpi%5E2%2830-2x%29%5E2%2Bx%5E2

y=cross%28pi%29%2A%22%221%2Fpi%5Ecross%282%29%2830-2x%29%5E2%2Bx%5E2

y=1%2Fpi%2830-2x%29%5E2%2Bx%5E2


[If you were taking calculus you would take the derivative here and set it
equal to 0, but I am assuming that you are taking college algebra, so you must
use the vertex formula.]

y=1%2Fpi%28900-120x%2B4x%5E2%29%2Bx%5E2

y=900%2Fpi-%28120%2Fpi%29x%2B%284%2Fpi%29x%5E2%2Bx%5E2

y=900%2Fpi-%28120%2Fpi%29x%2B%284%2Fpi%2B1%29x%5E2

or in descending powers:

y+=+%284%2Fpi%2B1%29x%5E2-%28120%2Fpi%29x%2B900%2Fpi

The coefficient of x%5E2 is positive so this represents
a parabola that opens upward, so its vertex will be at a minimum

To find the x-cordinate of the vertex, we use the vertex formula

x-coordinate of vertex = -b%2F%282a%29

x-coordinate of vertex = -%28%28-120%29%2Fpi%29%2F%282%284%2Fpi%2B1%29%29%22%22=%22%22%28%28120%29%2Fpi%29%2F%282%284%2Fpi%2B1%29%29%22%22=%22%22%28%28120%29%2Fpi%29%22%F7%22%282%284%2Fpi%2B1%29%29%22%22=%22%22%28%28120%29%2Fpi%29%22%F7%22%282%28%284%2Bpi%29%2Fpi%29%29%22%22=%22%22%28%28120%29%2Fpi%29%22%F7%22%28%282%284%2Bpi%29%29%2Fpi%29%29
%22%22=%22%22%28%28120%29%2Fpi%29%22%D7%22%28pi%2F%282%284%2Bpi%29%29%29%29%22%22=%22%22%28%28120%29%2Fcross%28pi%29%29%22%D7%22%28cross%28pi%29%2F%282%284%2Bpi%29%29%29%29%22%22=%22%22%22%D7%22%28120%2F%282%284%2Bpi%29%29%29%29%22%22=%22%2260%2F%284%2Bpi%29

So for the minimum area, the side of a square will be 60%2F%284%2Bpi%29 cm.
That is approximately 8.401487303 cm.

We will need to cut the wire at 4 times the side of the square.

So we must cut the wire at 4%22%D7%2260%2F%284%2Bpi%29 or 240%2F%284%2Bpi%29 
or about 33.60594921 cm from one end and, subtracting from 60,
26.39405079 cm from the other end.

Now for the maximum area. The problem is only defined for 0+%3C=+x+%3C=+15
When x=0, the square shrinks to 0 and the whole 60cm wire is made into a
circle.  When x=15, making the perimeter of the square 60 cm, the circle
shrinks to 0 and the whole 60cm wire is made into a square.  Since the parabola
opens upward, the maximum value is at one endpoint of the interval, either when
x=0 or when x=15.  It is well known that if a piece of wire is bent into a
circle or a square, the circle will have more area, so we could just assume the
maximum area would be when we "cut" the wire 0, or no, centimeters from the
end, and bend the whole wire into a circle. That is we don't cut the wire at
all.

Edwin