Question

A piece of wire 28 ft long is cut into two pieces. One piece is used to form a square, and the remaining piece is used to form a circle. Where should the wire be cut so that the combined area of the two figures is a maximum? (Round your answers to two decimal places.)

Answer 1

Answer:

Wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

Step-by-step explanation:

Length of the wire = 28 ft has been cut in two pieces.

One piece is used to form a square and remaining piece to form a circle.

Let the length of the wire which forms the square is 'l' ft.

Area of the square = (side)²

Perimeter of the square = 4(side) = l

Length of one side = $\frac{l}{4}$

So, the area of the square = $\frac{l^{2}}{16}$ ft²

Now length of the remaining part = perimeter of the circle = (28 - l) ft

2πr = (28 - l)

r = $\frac{28-l}{2\pi }$

Area of the circle formed = πr²

= $\frac{\pi(28-l)^{2} }{4(\pi )^{2} }$

= $\frac{(28-l)^{2}}{4\pi }$

Combined area of the square and circle

= $\frac{l^{2}}{16}$ + $\frac{(28-l)^{2}}{4\pi }$

= $\frac{l^{2}}{16}$ + $\frac{(28-l)^{2}}{4\pi }$

Now to maximize the area we will find the derivative of the area with respect to l.

$\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}]$

= $\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }]$

= $\frac{2l}{16}+\frac{(2l-56)}{4\pi }$

Now equate the derivative to zero.

$\frac{2l}{16}+\frac{(2l-56)}{4\pi }$ = 0

$(2l-56)=-\frac{2l}{16}\times 4\pi$

$(2l-56)=-\frac{\pi l}{2}$

$2l+\frac{\pi l}{2}=56$

$\frac{(4l+l\pi )}{2}=56$

l(4 + π) = 112

l(4 + 3.14) = 112

l = $\frac{112}{7.14}$

l = 15.69 ft

Length of the other part = 28 - 15.69 = 12.31 ft

Therefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft