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: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 = [tex]\frac{l}{4}[/tex]So, the area of the square = [tex]\frac{l^{2}}{16}[/tex] ft²Now length of the remaining part = perimeter of the circle = (28 - l) ft2πr = (28 - l)r = [tex]\frac{28-l}{2\pi }[/tex]Area of the circle formed = πr²= [tex]\frac{\pi(28-l)^{2} }{4(\pi )^{2} }[/tex]= [tex]\frac{(28-l)^{2}}{4\pi }[/tex]Combined area of the square and circle = [tex]\frac{l^{2}}{16}[/tex] + [tex]\frac{(28-l)^{2}}{4\pi }[/tex]= [tex]\frac{l^{2}}{16}[/tex] + [tex]\frac{(28-l)^{2}}{4\pi }[/tex]Now to maximize the area we will find the derivative of the area with respect to l.[tex]\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}][/tex]= [tex]\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }][/tex]= [tex]\frac{2l}{16}+\frac{(2l-56)}{4\pi }[/tex]Now equate the derivative to zero.[tex]\frac{2l}{16}+\frac{(2l-56)}{4\pi }[/tex] = 0[tex](2l-56)=-\frac{2l}{16}\times 4\pi[/tex][tex](2l-56)=-\frac{\pi l}{2}[/tex][tex]2l+\frac{\pi l}{2}=56[/tex][tex]\frac{(4l+l\pi )}{2}=56[/tex]l(4 + π) = 112l(4 + 3.14) = 112l = [tex]\frac{112}{7.14}[/tex]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