Finding maximum volume (where $x^2+y^2+z^2=r^2, x,y,z>0$)

Question

I would appreciate if somebody could help me with the following problem

Q. Finding maximum volume

$$(x+y+z)^2(xyz)^2 ~~~(\text{where} ~~x^2+y^2+z^2=r^2, x,y,z>0)$$

Answer 1

Hints: $2xy+2yz+2xz \le (x^2+y^2)+(y^2+z^2)+(x^2+z^2)$ and $\sqrt[3]{x^2y^2z^2} \le \dfrac{x^2+y^2+z^2}{3}$.

Answer 2

Note that the cost is zero if any of $x,y,z$ is zero, so we can relax the problem to $\max \{ ((x+y+z)xyz)^2 |\ x,y,z \ge 0, \ x^2+y^2+z^2 = r^2 \}$.

Since the variables are non-negative, we can simplify the cost to the equivalent $\max \{ (x+y+z)xyz |\ x,y,z \ge 0, \ x^2+y^2+z^2 = r^2 \}$.

The cost is continuous and the domain is compact, so we know a maximum exists. Furthermore, we know that the constraints $x,y,z \ge 0$ are not active at a solution. Hence the gradient corresponding to the constraint $x^2+y^2+z^2 = r^2$ will be non-zero and we can use Lagrange multipliers.

This gives $yz(2x+y+z) + \lambda 2 x = yz(x+y+z) + xyz+\lambda 2 x = 0$, and similarly for the $y,z$ components. Letting $\alpha = x+y+z >0$ and $C = xyz > 0$, we have $ \alpha yz +C + \lambda 2x = 0$, and similarly for $y,z$ components. In particular, we note that we must have $\lambda < 0$.

Now take two of these equations $\alpha yz +C + \lambda 2x = 0$ and $ \alpha xz +C + \lambda 2y = 0$, and subtract to give $ (\alpha z- 2 \lambda)(y-x) = 0$. Since $\alpha, z, -\lambda $ are strictly positive, we see that $x=y$, and similarly we have $z=x$. The constraint gives $x=y=z = \frac{1}{\sqrt{3}} r$.

Hence the maximum value of the original problem is $\frac{r^8}{9}$.