What are the extrema and saddle points of f(x,y) = xy + 1/x^3 + 1/y^2f(x,y)=xy+1x3+1y2?

1 Answer
Nov 13, 2015

The point (x,y)=((27/2)^(1/11),3 * (2/27)^{4/11}) approx (1.26694,1.16437)(x,y)=((272)111,3(227)411)(1.26694,1.16437) is a local minimum point.

Explanation:

The first-order partial derivatives are (partial f)/(partial x)=y-3x^{-4}fx=y3x4 and (partial f)/(partial y)=x-2y^{-3}fy=x2y3. Setting these both equal to zero results in the system y=3/x^(4)y=3x4 and x=2/y^{3}x=2y3. Subtituting the first equation into the second gives x=2/((3/x^{4})^3)=(2x^{12})/27x=2(3x4)3=2x1227. Since x !=0x0 in the domain of ff, this results in x^{11}=27/2x11=272 and x=(27/2)^{1/11}x=(272)111 so that y=3/((27/2)^{4/11})=3*(2/27)^{4/11}y=3(272)411=3(227)411

The second-order partial derivatives are (partial^{2} f)/(partial x^{2})=12x^{-5}2fx2=12x5, (partial^{2} f)/(partial y^{2})=6y^{-4}2fy2=6y4, and (partial^{2} f)/(partial x partial y)=(partial^{2} f)/(partial y partial x)=12fxy=2fyx=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}-1D=2fx22fy2(2fxy)2=72x5y41. 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.