Question

find the critical points of the function x^3+2y^3-xy and analyze them using the second-derivative test.

Answer 1

$$f(x,y)=x^3+2y^3-xy\\f_x(x,y)=3x^2-y\\f_y(x,y)=6y^2-x$$Now solve for both equal to \(0\):$$3x^2-y=0\\3x^2=y\\6y^2=x\\6(3x^2)^2=x\\6(9x^4)=x\\54x^4-x=0\\x(54x^3-1)=0\\54x^3=1\\x^3=\frac1{54}=\frac12\cdot\frac1{27}\\x=\frac1{3\sqrt[3]{2}}$$hence we have critical points at \(x=0,x=\dfrac1{3\sqrt[3]2}\)

Answer 2

the second-derivative test involves taking the determinant of the Hessian matrix:$$\mathbf{H}=\begin{bmatrix}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial y\partial x}\\\frac{\partial^2 f}{\partial x\partial y}&\frac{\partial^2 f}{\partial y^2}\end{bmatrix}\\\det\mathbf{H}=\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}-\frac{\partial^2 f}{\partial y\partial x}\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}-\left(\frac{\partial^2 f}{\partial y\partial x}\right)^2$$

Answer 3

$$f(x,y)=x^3+2y^3-xy\\f_x(x,y)=3x^2-y\\f_y(x,y)=6y^2-x\\f_{xx}(x,y)=6x\\f_{xy}(x,y)=-1\\f_{yy}(x,y)=12y$$hence we have $$\det\mathbf{H}=(6x)(12y)-(-1)^2=72xy-1$$Evaluating at our critical points we have:$$(0,0)\colon \det\mathbf{H}=-1<0\\\left(\frac1{3\sqrt[3]2},\frac1{3\sqrt[3]4}\right)\colon\det\mathbf{H}=4-1=3>0$$Observe that at \((0,0)\) it follows we have a saddle point, whereas at our other point we have an extremum. To determine which kind, observe we have \(f_{xx}>0,f_{yy}>0\) i.e. we have positive curvature at the point, and thus this critical point is a local minimum.

Answer 4

@FutureMathProfessor well then we'd have \(\det\mathbf{H}<0\) i.e. a saddle point -- does that make sense? :-) we'd have positive curvature parallel to the x-axis but negative curvature parallel to the y-axis e.g.