Question

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

Answer 1

The point `(x,y)=((27/2)^(1/11),3 * (2/27)^{4/11}) approx (1.26694,1.16437)` is a local minimum point.

Explanation:

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

The second-order partial derivatives are `(partial^{2} f)/(partial x^{2})=12x^{-5}`, `(partial^{2} f)/(partial y^{2})=6y^{-4}`, and `(partial^{2} f)/(partial x partial y)=(partial^{2} f)/(partial y partial x)=1`.

The discriminant is therefore `D=(partial^{2} f)/(partial x^{2})*(partial^{2} f)/(partial y^{2})-((partial^{2} f)/(partial x partial y))^{2}=72x^{-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.