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 minimum?

1 Answer
Mar 20, 2018

The length of wire required for the square will be #[80sqrt3]/[72+32sqrt3# and the length of wire required for the triangle will be #10 - [80sqrt3]/[72+32sqrt3#. [ Answers in m]

Explanation:

Let the side of the square have a length #x# m and the side of the triangle have a length of #y # m.

The combined lengths of the square and rectangle will equal #4x+3y # and this is equal to #10# [given].

#4x+3y=10#, therefore, #y=[10-4x]/[3]#..............#[1]#.
Area of equilateral triangle of side length #y# =#sqrt3/4 y^2#. [This derived from Pythagoras theorem].
The total area of both square and triangle =#x^2 + [sqrt3/4][[10-4x]/3]^2#..............#[2]# Since #y= [10-4x]/3# from ......#[1]#

So differentiating ..#[2]# with respect to A [area] to find a turning point and hence a max or min, using the chain rule .

#dA/dx# = #72x-80sqrt3+32sqrt3x=0# [ for max/min] ,and after manipulation results in #x=[80sqrt3]/[72+32sqrt3#

#d^2a/dx^2# = #72+32sqrt3#, which is positive [ whatever the value of x] and so by the second derivative test , the value of #x# obtained from the first derivative will minimise the area function.

Hope this helps.