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?
 A: 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.
A: Since $x^2y=15$, we have $y\gt0$. To maximize $x+y$, we want $x\gt0$ as well. However, for $x\gt0$, we have the reverse inequality
$$
\frac{x+y}3=\overbrace{\frac{2\cdot\color{#C00}{\frac x2}+\color{#090}{y}}3\ge\left(\left(\color{#C00}{\frac x2}\right)^2\color{#090}{y}\right)^{1/3}}^\text{AM-GM}=\left(\frac{15}4\right)^{1/3}
$$
But as Alan says, we have $y=\frac{15}{x^2}$, so
$$
x+y=x+\frac{15}{x^2}
$$
which can be made as large as we wish by making $x$ very small or very big.

If we consider $x\lt0$, then $x+\frac{15}{x^2}:(-\infty,0)\to(-\infty,\infty)$, so there is no minimum or maximum.
A: Lagrange method only finds LOCAL maxima and minima. Then you need what happens in the boundary of the domain. In this case the function has only a local minimum at $(\sqrt[3]{30},\tfrac{1}{2}\sqrt[3]{30})$, but it has no global minimum nor maximum.
