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. How should the wire be cut so that the total area enclosed is maximum?

1 Answer
Jun 28, 2018

The triangle should have #3.022# metres of wire, which means the square should have 6.978 meters

Explanation:

We know the perimeter of the two shapes combined will be the amount of wire, 10 m.

Call the two pieces #a# and #b#. Then #P =10= a+b#. Now we need to find an expression for area.

#A_“square” = (a/4)^2#
#A_“triangle” = (b/3)sqrt(b^2 - (b/2)^2)(1/2)#

Since #a+b=10#, #a= 10-b#

#A_“square” = ((10 -b)/4)^2#

#A_“combined” = ((10-b)/4)^2+ b^2/12sqrt(3)#

#A_“combined” = (100 - 20b +b^2)/16 + b^2/12sqrt(3)#

#A’_“combined” = -20/16+ b/8 + b/6sqrt(3)#

Look for critical values.

#0= -20/16 + (3b + 4bsqrt(3))/24#

#20/16(24) =b(3+4sqrt(3))#

#b = 30/(3 +4sqrt(3))~~3.022# m

It follows that #a=10-30/(3+4sqrt(3)) ~~6.978#.

Hopefully this helps!