# What are the extrema and saddle points of f(x,y) = xy + 1/x^3 + 1/y^2?

Nov 13, 2015

The point $\left(x , y\right) = \left({\left(\frac{27}{2}\right)}^{\frac{1}{11}} , 3 \cdot {\left(\frac{2}{27}\right)}^{\frac{4}{11}}\right) \approx \left(1.26694 , 1.16437\right)$ is a local minimum point.

#### Explanation:

The first-order partial derivatives are $\frac{\partial f}{\partial x} = y - 3 {x}^{- 4}$ and $\frac{\partial f}{\partial y} = x - 2 {y}^{- 3}$. Setting these both equal to zero results in the system $y = \frac{3}{x} ^ \left(4\right)$ and $x = \frac{2}{y} ^ \left\{3\right\}$. Subtituting the first equation into the second gives $x = \frac{2}{{\left(\frac{3}{x} ^ \left\{4\right\}\right)}^{3}} = \frac{2 {x}^{12}}{27}$. Since $x \ne 0$ in the domain of $f$, this results in ${x}^{11} = \frac{27}{2}$ and $x = {\left(\frac{27}{2}\right)}^{\frac{1}{11}}$ so that $y = \frac{3}{{\left(\frac{27}{2}\right)}^{\frac{4}{11}}} = 3 \cdot {\left(\frac{2}{27}\right)}^{\frac{4}{11}}$

The second-order partial derivatives are $\frac{{\partial}^{2} f}{\partial {x}^{2}} = 12 {x}^{- 5}$, $\frac{{\partial}^{2} f}{\partial {y}^{2}} = 6 {y}^{- 4}$, and $\frac{{\partial}^{2} f}{\partial x \partial y} = \frac{{\partial}^{2} f}{\partial y \partial x} = 1$.

The discriminant is therefore $D = \frac{{\partial}^{2} f}{\partial {x}^{2}} \cdot \frac{{\partial}^{2} f}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}^{2} = 72 {x}^{- 5} {y}^{- 4} - 1$. This is positive at the critical point.

Since the pure (non-mixed) second-order partial derivatives are also positive, it follows that the critical point is a local minimum.