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

Answer 1

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.