MATH100 December 2018
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A one metre long piece of wire is cut into two pieces and one piece is bent into a square, while the other
is bent into an equilateral triangle. Where, if anywhere, should the wire be cut so that the total area enclosed in the square and triangle is minimal?
Where, if anywhere, should the wire be cut so that the total area enclosed in the square and triangle is maximal?
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Suppose that the wire is cut into a piece of length metres and one of length metres, where . Further, suppose that the metre long piece is used to make a square and that the metre long piece is used to make a triangle.
What would the total enclosed area be in terms of ? (This is what must be optimized.)
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Let us suppose that we cut the wire into pieces, one of length metres, and one of length metres, where we suppose that . We use the metre long piece to make a square and the metre long piece to make a triangle.
The square has side lengths metres, while the triangle has side lengths metres. It follows that the area of the square is square metres, while the area of the triangle is square metres. The total area enclosed by our wire is thus square metres.
Now and so has no singular points and only one critical point, . Notice that this value lies in the interval and that for , while for . We deduce that has a global minimum (for ) at , and a global maximum at either or (each of which corresponds to not cutting the wire at all).
We calculate and . Since , we have that and hence has a global maximum at (which corresponds to not cutting the wire and instead using all of it to make a square).
In other words, the wire should be cut along its length to obtain the minimal area, and should not be cut but use the entire wire to bent into a square to obtain the maximal area.
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