Question

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 52 inches long and cuts it into two pieces. Steve takes the first piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. What should the lengths of the wires be so that the total area of the circle and square combined is as small as possible

Answer 1

Steve should use:

• 8.53 Inch of Wire to make the circle
• 43.47 Inches of Wire to make the Square.

Explanation:

Let Steve cut the wire so that the first piece has length x.

Therefore, the second piece will have a length of (52 - x).

The wire of length x is used to make a circle

Circumference of a Circle,

`c = 2\pi r\\Therefore\\2\pi r=x\\r=(x)/(2\pi)`

`\text{Area of a circle, A}=\pi r^2= \pi ((x)/(2\pi))^2=\pi ((x^2)/(4\pi^2))\\A=(x^2)/(4\pi)`

The wire of length (52-x) is used to make a square.

`\text{Side length of the Square,}`
`s= (52-x)/(4)`

`\text{Area of the Square},s^2= \left((52-x)/(4)\right)^2=((52-x)^2)/(16)=(x^2-104x+2704)/(16)`

Total Area = Area of Circle + Area of Square

`Area=(x^2)/(4\pi)+(x^2-104x+2704)/(16)`

Let us simplify the expression

`Area=(16x^2+\pi(x^2-104x+2704))/(16\pi)\\=(16x^2+\pi x^2-104\pix+2704\pi)/(16\pi)\\=(x^2(16+\pi)-104\pi x+2704\pi)/(16\pi)\\=(x^2(16+\pi))/(16\pi)-(104\pi x)/(16\pi)+(2704\pi)/(16\pi)\\A=(x^2(16+\pi))/(16\pi)-6.5x+169`

This is the function of a parabola which opens up.

To find where A is minimum, find the axis of symmetry.

`$Using \: x=-(b)/(2a)`

`a=(16+\pi)/(16\pi), b=-6.5\\ x=-(-6.5)/(2((16+\pi)/(16\pi)))=8.53\:Inches`

Steve should cut the wire so that the length of wire used to make a circle is 8.53 Inches.

Length of wire used to make the square =52-8.53=43.47 Inches