A wire 9 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each: For the equilateral triangle:

cutting point to minimize total area = 3.39 m

Step-by-step explanation:

Length of wire = 9m

wire is cut into 2 i.e. length of new wires = 4.5 m each

x = 0 ( where the wire will be cut )

Determine the length of wire that will yield minimum total area of figures

i) For equilateral triangle

assume side of triangle = b

length of wire = 9 m

i.e. 3b = 9m  hence :  b = 9/3 = 3m

 area of equilateral triangle = b^2  =  3.89 m^2

when there is no cut the value of x = 9 m

∴ perimeter of triangle = 3b = 9 - x ,   b = 9 - x / 3

ii) let the radius of circle = c

2πr = 9 ∴ r = 9 / 2π

hence area of circle = πr^2 = 81 / 4π = 6.44 m^2

perimeter of a circle:  2πr = x

                                     ∴ r = x / 2π

attached below is the remaining part of the solution