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

Answer:

Steve should use:

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

Step-by-step 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,

$c = 2\pi r\\Therefore\\2\pi r=x\\r=\dfrac{x}{2\pi}$

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

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

$\text{Side length of the Square,}$ $s= \dfrac{52-x}{4}$

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

Total Area = Area of Circle + Area of Square

$Area=\dfrac{x^2}{4\pi}+\dfrac{x^2-104x+2704}{16}$

Let us simplify the expression

$Area=\dfrac{16x^2+\pi(x^2-104x+2704)}{16\pi}\\=\dfrac{16x^2+\pi x^2-104\pix+2704\pi}{16\pi}\\=\dfrac{x^2(16+\pi)-104\pi x+2704\pi}{16\pi}\\=\dfrac{x^2(16+\pi)}{16\pi}-\dfrac{104\pi x}{16\pi}+\dfrac{2704\pi}{16\pi}\\A=\dfrac{x^2(16+\pi)}{16\pi}-6.5x+169$

This is the function of a parabola which opens up.

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

$Using \: x=-\frac{b}{2a}$

$a=\dfrac{16+\pi}{16\pi}, b=-6.5\\ x=-\dfrac{-6.5}{2(\dfrac{16+\pi}{16\pi})}=8.53\:Inches$

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