TITLE: Find two positive real numbers, whose difference is 100 and whose product is a minimum
QUESTION [11 upvotes]: First off, this is a single-variable calculus optimization problem. At first glance, the problem seemed extremely trivial, however the solution to it seems to be deceptively tricky (at least to me at this moment in time).

Problem: Find two positive numbers whose difference is $100$ and whose product is a minimum


My Attempted Solution :
Let's assume, $a, b \in \mathbb{R^+}$, $b < a$ 
First we set up an equation for the difference of $a, b$, I've used absolute values to restrict $a,b$ to values $\geq 0$
$$|a| - |b| = 100$$
$$\implies |a| = 100 + |b|$$
Next we set up an equation for the product of $a, b$
$$(|a|)(|b|) = b^2 +  100|b| \ \ \ \ $$
Defining a function $f$, to minimize to product of $a , b$ :
$$f(b) = b^2 + 100|b|$$
$$\implies 
    f'(b) = \begin{cases}
2b +100, \ \ \ \text{if} \ \ b \geq 0 \ \ \ \ (1)\\
2b - 100, \ \ \ \text{if} \ \ b < 0 \  \ \ \ (2)\\
\end{cases}  $$
Solving for $f'(b) =0$, with case $(1)$, yields $b = -50$, an immediate contradiction. Solving for $f'(b) =0$, with case $(2)$, gives $b=50$, $a=150$. Although that is a valid solution (EDIT: As correctly pointed out in an answer below, this is also a contradiction), they are not the minimum values.
The correct values, just thinking about it should be, $a = 100$, $b = 0$. However, trying to minimize it, using the equations I set up, don't yield the correct solution. Why is that so?. I don't seem to have made any mistakes as far as I can tell

REPLY [12 votes]: Without loss of generality assume $b > a$ so $a + 100 = b$.
The product is $ab = a(a+100) = a^2 + 100a$.
To find an extrema we must solve for $f(x) = x^2 + 100x$ $f'(x) = 0$.
That is $2x + 100 = 0$ so $x = -50$ is the only extremum.
To see what type of extremum it is we must evaluate $f''(x)$ is at $x = -50$.  $f''(x) = 2$ so $f''(50) = 2 > 0$.  $x = 50$ it is a minimum.
So $b = 50$ and $a = -50$ is the minimum product. But those aren't positive.
We need to find the minimum positive product.  If $0 < x$ then $f'(x) > 2*0 + 100$ so the product is increasing.
So the minimum product occurs at $a = \min (0, \infty)$ and $b = \min (100, \infty)$.  
But .... there are no such real numbers.
Which... isn't a problem.  You fell for the oldest trick in the book-- one which every mathematician I have ever known has fallen for once or twice--- that just because a problem might ask for something, that doesn't mean the thing exists.
Now had it be two non-negative real numbers the question would have been valid, but it wasn't and it isn't.