<s>[INST] 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?
 [/INST] 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.
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