It is obvious that we are doing too much arithmetic and algebra in our schools. This seems to be true all over the world. Only very few will have a need for algebraic manipulations („search x“) in their career. Moreover, these things can easily be done by computers now, much more accurate and less error prone. Plus, you can handle far more complex manipulations.

Let me give an example. If you ask for the can, which has the least surface and contains 200ml, you can handle this with algebra and calculus. You end with an equation of degree 3, with the solution

\(r = \left(\dfrac{100}{\pi}\right)^{1/3}\)

Let me do it on a computer.

>V &= pi*r^2*h; O &= 2*pi*r^2+2*pi*r*h; >&solve(V=200,h); fr &= O with % 2 400 2 pi r + --- r >&solve(diff(fr,r))[3] 1/3 100 r = ------ 1/3 pi

In fact, h=2r, which is not a good looking can. Indeed, you were able to solve this algebraically. It gets a bit easier, if you use Lagrange’s method, but that is another story.

On this page, Karl Fisch asked if it is really worth the effort. After all, the student has to learn quite a bit of calculus to solve a problem, which can be solved with numerical methods easily.

For me, this is a borderline case. The modelling is certainly nice and involves the geometry you need in everyday life. The algebraic manipulations necessary to derive the equation, which should be minized are just the maximal ones I would expect from any high school student. The calculus is also on elementary level. However, such problems could also be solved on medium level, where no calculus is available. Then the best thing to is to use a computer and a numerical system.

Clearly, most real life problems are impossible to solve exactly. My favorite example is a string around a sphere (say the earth), which is 1m longer than the perimeter of the sphere. But the string should be lifted at one point maximally. You will have to solve tan(x)-x=c, which can only be done numerically or approximately.

There is an interesting point here. If the ball is very large, c is very small, and it might be interesting to know the approximative behavior of x(c) as c goes to 0. To derive this, you can develop tan(x) for x, and get an approximation

\(\dfrac{x^3}{3} \approx c\)

While it is possible, deriving the proper asymptotics for x(c) numerically is a bit more difficult.