Question

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Round your answers to two decimal places.)

1. minhkhue
The maximum and the minimum total area inclosed are 6.25m² and 2.72m²
What is an area?
⇒ The area is the region bounded by the shape of an object.
Calculation:
Let x is the length of the piece of wire for the equilateral triangle, whose side will then be (x/3) m long.
⇒  (10 – x) = length of the remaining piece for the square, whose side will then be (10 – x)/4 m long.
h = height of the equilateral triangle = h = a×√3/2 = (x/3)×(√3/2)
(Where a is the side of the triangle)
A(x) = Total area of the square formed + Total area of the triangle formed
Then    A(x) = [(10 – x)/4]²+ (1/2)(x/3)(x/6)√3
= [(100 – 20x + x²)/16] +  (√3/36)x² .

The maximum total area enclosed :
⇒ The entire 10m length should be allocated to the square because a square produces more area per unit of its perimeter than does a triangle.
Thus if all square, then x=0 and A(0)  = [2.5]² = 6.25 m² = Maximumarea
If all triangle, then x = 10  and  A(10) = (1.732/36)(100) = 4.811 m² .
So, the maximum area occurs when it’s all used to make a square of side 2.5 m.
The minimum total area enclosed is :
⇒ We want a relatively small square and a small triangle.
We are going to Find x by setting the derivative of A(x) to zero.
d[ A(x)]/dx  =   [(-20 + 2x)/16] + (2√3/36)x = 0
=  -5/4  + (1/8)x  +  (√3/18)x  = 0
x =  5/[4 (1/8 + √3/18)]  = 5/[ 4(0.125 + 0.09622)] = 5/(0.88489)  ⇒5.65 m perimeter of the triangle

and (10 – x) = 4.35 = perimeter of square
And
A(5.65) = [4.35/4]² +  (√3/36)(5.65)²
= 1.1827 from the square+1.5358 from the triangle
=  2.719m² total area = Minimum area
⇒ On rounding off it will be 2.72m²
Hence the maximum and the minimum areas are 6.25m² and 2.72m²