Maximum of $x^3+y^3+z^3$ with $x+y+z=3$

Question

It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.

My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,

$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).

So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.

Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated

Answer 1

Hint : Use :

$$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$

Answer 2

Here is how I would solve it, First, I would eliminate x. \begin{eqnarray*} x &=& 3 - y - z \\ f(x,y,z) &=& (3 - y - z)^3 + y^3 + z^3 \end{eqnarray*} then I would apply the second derivative test. More information about the second derivative test can be found at the following URL:
http://faculty.csuci.edu/brian.sittinger/2nd_DerivTest.pdf

First, we find the partial derivatives: \begin{eqnarray*} f_y &=& -3(3 - y - z)^2 + 3y^2 f_z &=& -3(3 - y - z)^2 + 3z^2 f_yz &=& 6(3 - y - z) + 6y \\ f_yy &=& -6(3 - y - z) + 6y \\ f_zz &=& -6(3 - y - z) + 6z \end{eqnarray*} Now, we find the critical points: \begin{eqnarray*} -3(3 - y - z)^2 + 3y^2 &=& 0 \\ -3(3 - y - z)^2 + 3z^2 &=& 0 \\ 3y^2 - 3z^2 &=& 0 \\ y^2 &=& z^2 \\ \end{eqnarray*} Therefore, $y = z = 0$ is a critical point. However, it is a minimum not a maximum. We also have an infinity number of critical points, so I am not sure how to proceed.

Bob

Answer 3

$$ \begin{eqnarray} &(x+y+z)^3 &=& x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)\\ \implies & x^3 + y^3 + z^3 &=& (x+y+z)^3 - 3(x+y)(y+z)(z+x)\\ && = & 27 - 3(x+y)(y+z)(z+x) \end{eqnarray} $$

Now, $x^3+y^3+z^3$ is maximum when $t = (x+y)(y+z)(z+x)$ is minimum. Now since $x$, $y$ and $z$ are each non-negative, therefore $t$ is non-negative. Also, $x,\,y,\,z \in [0,\,2]$. So, $t$ takes minimum value when the variables take values $0,\,1,\,2$. So, $t_\text{min} = (0+1)(1+2)(2+0) = 6$.

So $\max (x^3+y^3+z^3)=27-3\times6=9$.

Answer 4

You have correctly established that $z=0$. From there you have $y=3-x$ so substitute that into $f$. As $y\le2$ then $1\le x\le2$.

$$f(x,3-x,0)=x^3+(3-x)^3=9x^2+27x+27=9(x^2+3x+3)$$

$$=9\left(x-\frac{3}{2}\right)^2+\frac{27}{4}$$

This quadratic has minimum at $x=\frac{3}{2}$ and the maximum is only limited by the domain of $x$ which leads to the answer of $x=1$ or $x=2$ so the three numbers are $0,1,2$ and the maximum is $9$.

Answer 5

Let $x\geq y\geq z$.

Since $f(x)=x^3$ is a convex function and $(2,1,0)\succ(x,y,z)$, by Karamata we obtain: $$x^3+y^3+z^3\leq2^3+1^3+0^3=9$$ Done!