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?
Best Answer
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.