Science:Math Exam Resources/Courses/MATH100/December 2018/Question 10

MATH100 December 2018

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Question 10
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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Hint
Suppose that the wire is cut into a piece of length $x$ metres and one of length $1-x$ metres, where $0\leq x\leq 1$ . Further, suppose that the $x$ metre long piece is used to make a square and that the $1-x$ metre long piece is used to make a triangle. What would the total enclosed area be in terms of $x$ ? (This is what must be optimized.)

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let us suppose that we cut the wire into $2$ pieces, one of length $x$ metres, and one of length $1-x$ metres, where we suppose that $0\leq x\leq 1$ . We use the $x$ metre long piece to make a square and the $1-x$ metre long piece to make a triangle. The square has side lengths $x/4$ metres, while the triangle has side lengths $(1-x)/3$ metres. It follows that the area of the square is $(x/4)^{2}=x^{2}/16$ square metres, while the area of the triangle is ${\frac {\sqrt {3}}{4}}\left({\frac {1-x}{3}}\right)^{2}={\frac {{\sqrt {3}}(1-x)^{2}}{36}}$ square metres. The total area enclosed by our wire is thus $A(x)={\frac {x^{2}}{16}}+{\frac {{\sqrt {3}}(1-x)^{2}}{36}}$ square metres. Now $A'(x)={\frac {x}{8}}-{\frac {{\sqrt {3}}(1-x)}{18}}={\frac {(9+4{\sqrt {3}})x-4{\sqrt {3}}}{72}}$ and so $A$ has no singular points and only one critical point, $x=x_{0}={\frac {4{\sqrt {3}}}{9+4{\sqrt {3}}}}$ . Notice that this value $x_{0}$ lies in the interval $(0,1)$ and that $A'(x)<0$ for $x\in (0,x_{0})$ , while $A'(x)>0$ for $x\in (x_{0},1)$ . We deduce that $A$ has a global minimum (for $x\in [0,1]$ ) at $x_{0}$ , and a global maximum at either $x=0$ or $x=1$ (each of which corresponds to not cutting the wire at all). We calculate $A(0)={\frac {\sqrt {3}}{36}}$ and $A(1)={\frac {1}{16}}$ . Since ${\sqrt {3}}<2$ , we have that ${\frac {\sqrt {3}}{36}}<{\frac {2}{36}}<{\frac {1}{16}}$ and hence $A$ has a global maximum at $x=1$ (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 $\color {blue}{\frac {4{\sqrt {3}}}{9+4{\sqrt {3}}}}{\text{ m}}$ 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.