How to find the max and min of a wire cut into a triangle and square

Question

A piece of wire 10m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. Find the maximum and minimum possible area that can be enclosed. by the wire.

Answer 1

Brittany,

 

Let          x = 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 = (x/6) tan 60^o = (x/6)√3.

          A(x) = Total area of the square plus the triangle formed from bending the 2 pieces of wire.

 

Then    A(x) = [(10 - x)/4]²                 +    (1/2)(x/3)(x/6)√3

                  = [(100 - 20x + x²)/16]    +    (√3/36)x² .

 

For MAXIMUM Total area:

 

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 = MAX area

If all triangle, then x = 10  and  A(10) = (1.732/36)(100) = 4.811 m² .

So, the maximum area occurs when its all used to make a square of side 2.5 m.

 

For MINIMUM Total area.

 

You want a relatively small square and a small triangle.  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 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 triangle  =  2.719 m² total area. = MIN area