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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?

user577215664
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user147485
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    Are you sure the problem requires both numbers to positive, rather than nonnegative? There isn't actually a maximum if both are required to be positive. – bzc Jan 29 '18 at 00:29
  • I agree! I am proofing someone's calculus work and I think there was a typo in the original question. I just wanted to make sure I didn't miss something obvious. Someone else suggested to define x>y then as x increases from 1/2 towards one, b would decrease from 1/2 towards 0. But that seems to not be mathematically sound. – user147485 Jan 29 '18 at 00:34
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    The substitution seems to flip the sign: from $x^3+y^3$ to $x^3-(..)^3$. – Sjoerd Jan 29 '18 at 01:04
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    A maximum of what? Maximum of $x^3 + y^3$? Okay, over what interval, if any? – jpmc26 Jan 29 '18 at 02:17
  • @user147485 you can say "without loss of generality" that $x \ge y$ because those are simply names. – muru Jan 29 '18 at 07:45
  • Shouldn't that be $x^3 +(1-x)^3 = max$? – Steven Alexis Gregory Jan 29 '18 at 10:39

6 Answers6

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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".

Weijun Zhou
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  • I would say it's not clear over what interval the maximum is. The question doesn't seem to specify. – jpmc26 Jan 29 '18 at 02:16
  • If $x$ and $y$ are nonnegative, then $x$ is in the closed interval $[0,1]$. Even if the question is unclear, the attempted solution made it clearer. – Weijun Zhou Jan 29 '18 at 02:18
  • Isn't an unqualified maximum usually understood to be over the maximum value over for any member of the domain (and codomain) of the function, though? $x^3 + y^3$ accepts many more values of $x$ and $y$ than just $[0, 1]$. – jpmc26 Jan 29 '18 at 02:19
  • The problem imposes restriction on the value that $x$ and $y$ can take. Without such restrictions it would be indeed as you say. – Weijun Zhou Jan 29 '18 at 02:23
  • Also, as dxiv points out below, the question restricts the values to $(0,1]$, not $[0,1]$. – jpmc26 Jan 29 '18 at 02:23
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    No. The original question restricts the values to $(0,1)$. So there is not a maximum. I stated that in my answer. – Weijun Zhou Jan 29 '18 at 02:25
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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 $\{a,b\}=\{0,1\}$.

dxiv
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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.

Ethan Bolker
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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.

Vassilis Markos
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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 $x=1$.

enter image description here

Steven Alexis Gregory
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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)

William Porter
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