Course:MATH110/Archive/2010-2011/003/Notes/Optimization/Problem 3

Question

A rope of length $L$ is to be cut in two parts. The first part is used to create an equilateral triangle and the second one to construct a square. Where should the rope be cut in order for the total area of the two shapes to be minimal? maximal?

Solution$\quad$

Before reading the solution, you should really try the problem for yourself. By should I mean that if you don't you will not gain any understanding, whatever you might believe, this is 100% guaranteed. So do yourself a favour, try the problem and look at the solution once you've gave it an authentic try. Solution$\quad$ Let us say that we cut the rope in two parts, the first having length $x$ and the second having length $y$. Clearly we have that $x+y=L$ so this means that if we know the value of $x$ we'll know immediately the value of $y$ and conversely. The first piece will be used to construct an equilateral triangle. Since the piece of rope we're using for this has length $x$ each side of the equilateral triangle will have a third of that length. To compute the area of the triangle, we need to know the length of its height. Consider the following picture which shows the equilateral triangle and its height, highlighting that one half of the figure is a right triangle. Using the Pythagorean theorem, we obtain that the length of the height of the triangle is: ${\sqrt {\left({\frac {x}{3}}\right)^{2}-\left({\frac {x}{6}}\right)^{2}}}={\sqrt {\frac {27x^{2}}{36}}}={\frac {x3{\sqrt {3}}}{6}}={\frac {\sqrt {3}}{2}}x$ This allows us to compute the area of the equilateral triangle to be ${\dfrac {1}{2}}{\frac {x}{3}}{\frac {\sqrt {3}}{2}}x$ (one half multiplied by the product of the length of the base and the length of the height) which can be simplified to be ${\dfrac {\sqrt {3}}{12}}x^{2}$ Now that we computed the area of the equilateral triangle, we can get to the area of the square (this will be much easier). Since we're using the rest of the rope which has length $y$ we obtain that each side of the square will be a quarter of that length and hence the area of the square will be given by $\left({\frac {y}{4}}\right)^{2}={\frac {y^{2}}{16}}$ Now we don't want to keep using both $x$ and $y$ at the same time, so since we know these two sum up to the total length of the rope, denoted by $L$ we can actually write $y=L-x$ and use this relation to write both the areas of the triangle and the square in terms of the variable $x$ only (of course we could do the opposite choice and write everything in terms of $y$, it wouldn't make the problem harder or simpler). We obtain that the combined are of the equilateral triangle and of the square is described by the following function. $A(x)={\dfrac {\sqrt {3}}{12}}x^{2}+{\frac {(L-x)^{2}}{16}}={\frac {\sqrt {3}}{12}}x^{2}+{\frac {L^{2}}{16}}-{\frac {2Lx}{16}}+{\frac {x^{2}}{16}}={\frac {4{\sqrt {3}}+3}{48}}x^{2}-{\frac {L}{8}}x+{\frac {L^{2}}{16}}$ We finally have a function describing the total area, we're ready to now use calculus to determine the extremums of this function. Before doing this, it's important to notice that we'll only study this function on the domain $[0,L]$ since the value of $x$ has to remain between 0 and $L$. Let's compute the derivative of this function. We obtain: $A'(x)={\frac {4{\sqrt {3}}+3}{24}}x-{\frac {L}{8}}$ This function will have a critical point where the derivative takes the value 0 which occurs at $x={\dfrac {3L}{4{\sqrt {3}}+3}}\approx 30\%L$ And we have to remember to keep $x=0$ and $x=L$ as critical points as well since they constitute the boundaries of the domain of the function. Since the second derivative of the function is always positive (it's a constant), we know that the critical point we found at $x={\dfrac {3L}{4{\sqrt {3}}+3}}$ is a local minimum. To determine the maximum and the absolute minimum and maximum we need to compute the value of the total area at each of the three critical points that we have and compare. We obtain: $A(0)={\dfrac {L^{2}}{16}}\approx 6.3\%L^{2}$ (this actually means we're only doing a square). $A(L)={\dfrac {{\sqrt {3}}L^{2}}{12}}\approx 14.4\%L^{2}$ (this actually means we're only doing an equilateral triangle). $A\left({\dfrac {3L}{4{\sqrt {3}}+3}}\right)={\frac {{\sqrt {3}}L^{2}}{4(4{\sqrt {3}}+3)}}\approx 4.4\%L^{2}$ (some little algebraic simplifications are required to get there, nothing too bad, try it for yourself). Which allows us to see that we get the largest total area when we only have an equilateral triangle and we get the smallest total area when we use a combination of both shapes (as pointed above, roughly 30% is used for the equilateral triangle and the rest on the square). This whole explanation might appear to be long but there are a few things to consider. First there is a long explanation for the geometric part which consists of finding the area of an equilateral triangle if we're only given the length of its side (or more precisely in this case its perimeter). This is just elementary geometry, the actual calculus part is much smaller. Then, the second observation is that we answered two problems at the same time. We found both the minimal and the maximal area that can be obtained with this construction.