Question

Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum.

Answer 1

if you call one \(x\) the other is \(100-x\)
the sum of the squares is
\[x^2+(100-x)^2\]
minimize that one

Answer 2

when you multiply out you get \[2 x^2-200 x+10000\] min will be at the vertex, when \[x=-\frac{b}{2a}\]

Answer 3

that is the dumb math teacher way to do it
the real way is to think
you have \(x+y=100\) which is the same as \(y+x=100\) and you want to minimize \(x^2+y^2\) which is the same as \(y^2+x^2\)
in other words, you cannot tell \(x\) and \(y\) apart
so the minimum must be where they are equal, namely at \(x=y=50\)