Q:

Find the critical points of the following function. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. f(x, y) = x^4 + y^4 + 32x - 4y + 5 What are the critical points? Select the correct choice below and fill in any answer boxes within your choice. The critical point(s) is/are. (Type an ordered pair Use a comma to separate answers as needed.) There are no critical points. Identify any local maxima. Select the correct choice below and fill in any answer boxes within your choice. There are local maxima at. (Type an ordered pair. Use a comma to separate answers as needed.) There are no local maxima. Identify any local minima. Select the correct choice below and fill in any answer boxes within your choice. There are local minima at. (Type an ordered pair. Use a comma to separate answers as needed.) There are no local minima. Identify any saddle points. Select the correct choice below and fill in any answer boxes within your choice. There are saddle points at. (Type an ordered pair. Use a comma to separate answers as needed.) There are no saddle points.

Accepted Solution

A:
[tex]f(x,y)=x^4+y^4+32x-4y+5[/tex]has critical points where the partial derivatives vanish:[tex]f_x=4x^3+32=0\implies x=-2[/tex][tex]f_y=4y^3-4=0\implies y=1[/tex][tex]f[/tex] has Hessian matrix[tex]H(x,y)=\begin{bmatrix}12x^2&0\\0&12y^2\end{bmatrix}[/tex]At the critical point, the Hessian determinant is[tex]\det H(-2,1)=\begin{vmatrix}48&0\\0&12\end{vmatrix}=576>0[/tex]and [tex]f_{xx}(-2,1)=48>0[/tex], which tells us (-2, 1) is a local minimum with a value of [tex]f(-2,1)=-46[/tex].