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

1 Answer
Nov 13, 2015

Answer:

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.