# Minimizing perimeter given rectangle's area for 10-years-olds

I was recently in touch with some person from Russia how is busy with books for Russian elementary schools, in particularly I learned that now they give elementary set theory for the 2nd grade (8-years-olds) together with multiplications of natural numbers (I was happy to know that students are given something abstract at such age, but must admit they won't use it until they enter the university).

My question elementary geometry rather than set theory, though. In a presentation for a book for the 4th grade (10-years-old) the following problem was described.

A rectangular garden has an area of 900 sq.m, and you want to to a fence around it. What is the minimal length of the fence you need?

So, given a rectangle with an area of 900 sq.m. students are asked to find the minimal possible value of a perimeter of such a rectangle. Regarding their geometrical knowledge: they studied how to find the area of a rectangle given lengths of its sides, and perhaps they also know the area formula for right triangles. Regarding their arithmetics, they know addition, multiplication, division with remainders - they don't know fractions for example.

The way I'd solve the problem is to find the minimum of $x + 900/x$ which is far far beyond what these students can do. Even an expression by itself would be unknown for them, they only start solving "equations" of the form $3x=6$. I thus wonder, what could be the way students are assumed to solve this problem.

Currently, I think of two possible "solutions". First, students can start choosing different pairs of sides: $(2,450)$, $(3,300)$ etc. until the moment they notice that passing by $(30,30)$ starts increasing the perimeter, and "guess" the right answer. This is not much of a solution, though, as students cannot justify why that $(30,30)$ is indeed the right answer.

Second, students could have been told explicitly that among all rectangles with the same area, the square has the minimal perimeter.

My question is: are there any elementary solutions for this problem that could be found by an extremely smart student, but without assuming any additional knowledge to what I've described?

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Assuming the sides are natural numbers, there are only a few possibilities to try, because $900 = 2^2\cdot 3^2 \cdot 5^2$ has only $27$ divisors (and there is no reason why one wouldn't be able to see a pattern). – dtldarek Nov 13 '13 at 15:46
@dtldarek: agree, that what I meant by saying that student can't prove the optimality of $(30,30)$. Perhaps, in the statement of the problem it was written that sides are assumed to be natural numbers - which didn't appear in the presentation so +1, however let us look whether there are other ideas. – Ilya Nov 13 '13 at 15:48

The fact that the rectangle with maximal area for a given perimeter also has minimal perimeter for a given area is both intuitive and easy to prove: if rectangle $R'$ had the same area as rectangle $R$ but smaller perimeter, you could scale up the sides of $R'$ to get $R''$ with the same perimeter as $R$ and larger area.

Now given a perimeter $P$, if the length is $P/4 + x$ the width is $P/4 - x$, and then the area is $(P/4 + x)(P/4 - x) = (P/4)^2 - x^2$, which obviously has its maximum at $x=0$.

If you don't like an algebraic proof, here's a geometric one. Compare a rectangle (say $a$ by $b$ with $a > b$) and a square with the same perimeter ($(a+b)/2$ by $(a+b)/2$). To go from the rectangle (blue and red regions below) to the square (blue and green), you add the green rectangle ($(a+b)/2$ by $(a-b)/2$) and remove the red rectangle ($(a-b)/2$ by $b$). It's clear that you're adding more area than you're subtracting.

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I somehow dislike the fact that it sounds as if the garden did already exist, which kind of contradicts the question of finding an optimal shape. But that's besides the point.

I could imagine the following chain of thought, though I haven't worked with children that age so I might be far wrong.

Start by turning the problem around: instead of trying to make the fence as short as it will go, try to make the rectangle as large as it will go while maintaining the length of the fence. You might want to fool around with a piece of string tied in a loop and wrapped around four fingers. In any case, you can readily see that for extremely narrow rectangles, the area is almost zero. You might consider the two degerate zero area rectangles as two ends of a continuum of possibilities. Both ends are bad. Furthermore, the thing is symmetric: whether you make your rectangle narrow and horizontal, or narrow and vertical, you have the same low area. Making the sides more equal will increase the area, as an example (computing $a\times b$ for two different partitions $a+b$ of the same number, which is a lot easier than your $x+\frac{900}x$) will quickly verify. So both ends of the continuum are bad, from there things get better, and everything is symmetric. Now it isn't too hard to conjecture that things might keep getting better until symmetry forces them to do otherwise. This is no proof, but sound mathematical intuition, which in my opinion is all that is asked for here. So now you “know” that for a fixed length of fence, the square has maximal area.

With that knowledge, go back to the original problem. We are looking for the shortest fence for that given area, so we are looking for a fence for which this is the maximal area. We know that maximal area to be that of a square, so we can try to guess the square root. Due to the devision-with-remainder stuff, they might know that two trailing zeros indicate a number which can be divided by 10 two times. So they might be able to measure area not in m² but instead in (10m)². In Germany there is a name for that unit, and according to Interwiki links so is there in Russia. If they did conversions between units before, they might spot this a bit faster than they would otherwise. Now finding the square root of 9 should be rather simple if they have memorized their multiplication tables. A little trial and error will get them there.

Keep in mind that all of this is hypothetical, and requires quite a bit of a detour from the minimization problem as originally formulated to the maximization problem that might be more accessible. It requires a good intuition about the scale invariance of the underlying geometry: it does not matter that you scale the fence length to some arbitrary value and try maximizing the area. It does not matter that you measure lengths in 10m instead of 1m increments.

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The English name for $100~\mathrm{m}^2$ is are, but that is maybe not widely known in the English-speaking world. – Jeppe Stig Nielsen Feb 3 '15 at 1:56

Assuming the sides are natural numbers, there are only a few possibilities to try, because $900=2^2 \cdot 3^2 \cdot 5^2$ has only 27 divisors (and there is no reason why one wouldn't be able to see a pattern).

If the question would be posed the other way around, i.e. "what is the largest area you can fence with rectangle of perimeter $120$", then you can calculate differences (similar to differentials, but without the limit), and see that $(x+1)∗(60−(x+1))−x∗(60−x)=59−2x$ is bigger than $0$ for $x<30$. Of course, this is hard, but still I would consider it a valid problem (especially for gifted children).

To complement the above and @RobertIsrael's answer, even if the child is uncomfortable with squares of variables, purely geometrical argument also suffices. Indeed, it's not that hard to see when the green rectangle is larger than the red one.

$\hspace{100pt}$

I hope this helps $\ddot\smile$

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