A piece of wire of length 6363 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?

a.

35.2792 cm from one end (The square)

And 27.7208 cm from the other end (The circle)

b. See (b) explanation below

Step-by-step explanation:

Given

Length of Wire ,= 63cm

Let L be the length of one side of the square

Circumference of a circle = 2πr

Perimeter of a square = 4L

a. To minimise

4L + 2πr = 63 make r the subject of formula

2πr = 63 - 4L

r = (63 - 4L)/2π

r = (31.5 - 2L)/π

Let X = Area of the Square. + Area of the circle

X = L² + πr²

Substitute (31.5 - 2L)/π for r

So,

X² = L² + π((31.5 - 2L)/π)²

X² = L² + π(31.5 - 2L)²/π²

X² = L² + (31.5 - 2L)²/π

X² = L² + (992.25 - 126L + 4L²)/π

X² = L² + 992.25/π - 126L/π +4L²/π Collect Like Terms

X² = 992.25/π - 126L/π + 4L²/π + L²

X² = 992.25/π - 126L/π (4/π + 1)L² Arrange in descending order of power

X² = (4/π + 1)L² - 126L/π + 992.25/π

The coefficient of L² 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

i.e

L = -b/2a

L = - (-126/π) / (2 * (4/π + 1)

L = (126/π) / ( 2 * (4 + π)/π)

L = (126/π) /( (8 + 2π)/π)

L = 126/π * π/(8 + 2π)

L = (126)/(8 + 2π)

L = 63/(4 + π)

So, for the minimum area, the side of a square will be 63/(4 + π)

= 8.8198 cm Approximated

We will need to cut the wire at 4 times the side of the square. (i.e. the four sides of the square)

I.e.

4 * (63/(4 + π)) cm

Or

35.2792 cm from one end.

Subtract this result from 63, we'll get the other end.

i.e. 63 - 35.2792

= 27.7208 cm from the other end

b. To maximize

Now for the maximum area.

The problem is only defined for 0 ≤ L ≤ 63/4 which gives

0 ≤ L ≤ 15.75

When L=0, the square shrinks to 0 and the whole 63 cm wire is made into a circle.

Similarly, when L =15.75 cm, the whole 63 cm wire is made into a square, the circle shrinks to 0.

Since the parabola opens upward, the maximum value is at one endpoint of the interval, either when

L=0 or when L = 15.75.

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 will assume that 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.

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.