The sum of two positive numbers is 1. The sum of their cubes is a maximum. What are the numbers? I set this up and end up finding the minimum (the two numbers would both be $1/2$).  To find a maximum value, I could reflect the functions and use $y^3-x^3$ but I still end up finding $1/2$ as the two numbers.  It does make sense that two two values to produce a max would be 0 and 1 but I can't figure out how to set up the problem from the start.  What I have is...

$x+y=1 \\
x^3 + y^3 = max$

Subsitution...

$x^3 - (1-x)^3 = max \\
x^3 - 1 + 3x - 3x^2 + x^3 = max \\
2x^3 - 3x^2 + 3x - 1 = max \\
6x^2 - 6x + 3 = 0 \\
x = 1/2 \text{ which makes }y = 1/2 \\
$

That is where my issue is.  How do I set it up to find the sum of the cubes to be a max?
For a function defined on a closed interval. The maximum and minimum are found where the derivative is zero, or at the boundaries of the interval.

You can take derivative, but you also need to check for the case $x=0$ and $x=1$. By the way your original question does not have an answer. I suppose you should mean "nonnegative".
Your answer gives the MINIMUM. As explained above, the diff is not zero at the ends of the range, where the answer is. So that answer is (1-€)x€. 

(€=epsilon)