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?
 A: 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".
A: Hint: $\;1=(a+b)^3 = a^3+b^3+3ab(a+b) \implies a^3+b^3 \le 1\,$.
However, with $\,a, b \gt 0\,$ strictly positive, the maximum does not exist, since for $\,a \to 0$ and $b=1-a \to 1\,$ the sum $a^3+b^3$ can get arbitrarily close to $1$, but not equal $1$. If you allow for non-negative $a,b \color{red}{\ge} 0$, instead, then the maximum is $1$ and is attained for $\{a,b\}=\{0,1\}$.
A: You should draw the graph of the cubic function you found for $x$ after your substitution. Then you'll see what's going on.
You can't find maxima or minima by blindly finding out where a derivative vanishes. You have to think about what is going on at the ends of the domain too. 
In this case there really is no maximum, because you're requiring positive $x$ and $y$ so can't use $0$. You can get as close to a maximum value of $1$ as you wish, though.
A: Well, at first:
$$x^3+y^3=(x+y)(x^2-xy+y^2)$$
Since you want to maximize this and $x+y=1$, we get that:
$$x^3+y^3=x^2-xy+y^2$$
Now, you can see the above as a quadratic with respect to $x$ - a parabola, and calculate:
$$\Delta=y^2-4y^2=-3y^2<0$$
So, this parabola attains its minimum value - it has a positive leading coefficient - at $x_0=\frac{y}{2}$ and it maximum value on the boundaries of the interval that $x$ belong to. In our case, since we have $x\in(0,1)$, it is clear that this parabola cannot have a maximum value. However, if we let - as proposed by many - $x,y\in[0,1]$, we see that the so wanted maximum value can be attained for:
$$x=0,y=1$$
or, symmetrically, for $$x=1,y=0$$
Alternatively, you can substitute $y=1-x$ in the above quadratic and get that:
$$x^3+y^3=x^2-x(1-x)+(1-x)^2=x^2-x+x^2+x^2-2x+1=3x^2-3x+1$$
which, again has a global minimum and attains its maximum value either on $0$ or $1$ - actually, exactly on both of them.
A: If $x\ge 0$ and $y\ge 0$ and $x+y=1$, then $y=1-x$ and $0 \le x \le 1$
So $x^3+y^3=x^3 + (1-x)^3 = 3x^2-3x+1$. This is a parabola with axis of symmetry, and minimum, at $x = \frac 12$. It follow that the max values will be at $x=0$ and $x=1$.

A: 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)
