A wire 8 meters long is cut into two pieces. One piece is bent into a equilatera

Question

A wire 8 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: For the circle: (for both, include units) Where should the wire be cut to maximize the total area? Again, give the length of wire used for each: For the equilateral triangle: For the circle: (for both, include units)

Explanation / Answer

Let one part is x and other part is (8 - x)
Let (8 - x) is used to make equilateral triangle and x is used to make a circle

So Perimeter of equilateral triangle is
3a = (8 - x)
a = (8 - x)/3....................sides of an equilateral triangle

Circumference of circle is
2pir = x
r = x/2pi................radius of circle

Total area, A = Area of equilateral triangle + Area of circle
A = 1/4 sqrt(3) * [(8 - x)/3]2 + pi * (x/2pi)2
A = 1/36 sqrt(3) * (8 - x)2 + x2/4pi
dA/dx = -1/18 sqrt(3) * (8 - x) + x/2pi = 0
1/18 sqrt(3) * (8 - x) = x/2pi
pi sqrt(3) (8 - x) = 9x
8pi * sqrt(3) - pi * sqrt(3) x = 9x
x = 8pi * sqrt(3) / [9 + pi * sqrt(3)]...............Length used for circle for minimizing area

8 - x = 8 - 8pi * sqrt(3) / [9 + pi * sqrt(3)]
8 - x = [72 + 8pi*sqrt(3) - 8pi*(sqrt(3)] / [9 + pi * sqrt(3)]
8 - x = 72 / [9 + pi * sqrt(3)]........................Length used for equilateral triangle for minimizing area

For maximizing just reverse the case by taking x for triangle and 8-x for circle

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