Maximize $x + y$ with constraint $y \cdot x^2=15$. As the title says, I am asked to maximize the sum of two numbers $x + y$ when $x^2 \cdot y = 15$.
The thing is... I tried substitution but got nowhere, it seems $h(x) = x + \frac{15}{x^2}$ only has a minimum value, not a maximum.
I've been starting to think there is an error in the excercise. I tried Lagrange multiplicators next, having:
\begin{align}
f(x,y) &= x+y\\
g(x,y) &= x^2\cdot y - 15 = 0
\end{align}
Then:
\begin{align}
\nabla f = \lambda \nabla g
\end{align}
that gives:
\begin{align}
2xy &= 1\\
2x^2 &= 1\\
x^2\cdot y &= 15
\end{align}
then we find
\begin{align}
2xy &= 2x^2\\
x(2y - x) &= 0\\
x = 0 &\lor x = 2y
\end{align}
Now, from $x^2 \cdot y = 15$ we know $x,y>0$, then we only have one solution $x=2y$, applying that to $g(x)$ we get $x^2 \cdot \frac{x}{2} = 15 \implies x = \sqrt[3]{30}, y = \frac{\sqrt[3]{30}}{2}$.
Then the only point that I can use from Lagrange is $(\sqrt[3]{30}, \frac{\sqrt[3]{30}}{2})$. But what guarantee do I have that said point is the maximum of the sum of $x + y$?. Doesn't Lagrange only find extremes? How do I check this result?
As I said earlier, if I graph $h(x) = x + \frac{15}{x^2}$ in $R^2$ I find that $x=\sqrt[3]{30}$ is the value in the $x$ axis where the minimum value of $h(x)$ is, so this only feeds my suspicions about the problem having an error and the point needed being actually a minimum instead of a maximum.
Can anyone shed some light into this?

There is definitely an error in this.  You can see that from the logical framework,  the product of two numbers being fixed means you can take one to 0 and the other to infinity on slider-scale.  Then the sum goes to infinity.
