Q:

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.)

Accepted Solution

A:
Answer:Wire should be cut in two parts with the length = 15.69 ft and 12.31 ftStep-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)²Perimeter of the square = 4(side) = lLength of one side = $$\frac{l}{4}$$So, the area of the square = $$\frac{l^{2}}{16}$$ ft²Now length of the remaining part = perimeter of the circle = (28 - l) ft2πr = (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) = 112l(4 + 3.14) = 112l = $$\frac{112}{7.14}$$l = 15.69 ftLength of the other part = 28 - 15.69 = 12.31 ftTherefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft