Find two positive real numbers, whose difference is 100 and whose product is a minimum 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
 A: The minimum does not exists. There exists a minimizing sequence:
$$
(a_n,b_n)=(1/n,100+1/n)
$$
A: If you are trying to find an optimum, you need first to verify that an optimum exists. In this instance, one does not: as $b$ gets closer and closer to $0$ from above, $b(100+b)$ gets closer and closer to $0$ from above.
A: To answer the question Why is that so, there is a mistake. Your case (1) yields an immediate contradiction, as you've noted, but so does your case (2). It yields $b=50$, which violates your assumption that $b<0$. So your equations do lead to the conclusion that the min cannot occur where $f'(b)=0$.
A: If the minimum of a function lies on the boundary of the region of parameters, then simple differentiation will not work.  You must also check the boundaries, which in this case includes the (true) solution $(0,100)$.
A trivial illustration:  Find the minimum value of $f(x) = x$ for $x \ge 0$.  Differentiation is of no help in finding the solution at $x = 0$.
A: 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.
