Question

Find and classify the critical points of $z = ({x}^{2} - 2x)( {y}^{2} - 4y)$Local Maximums: Local Minimums: Saddle Points: For each classification, enter a list of ordered pairs (x,y) where the max/min/saddle occurs. If there are no points for a classification, enter DNE.

Answer 1

$z=(x^2-2x)(y^2-4y)$$\begin{gathered} z_x=(2x-2)(y^2-4y) \\ z_y=(x^2-2x)(2y-4) \end{gathered}$

We also need the second derivatives;

$\begin{gathered} z_{xx}=2(y^2-4y) \\ z_{yy}=2(x^2-2x) \\ z_{xy}=(2x-2)(2y-4) \end{gathered}$

Equate the first derivative to zero;

$Critical\text{ point is \lparen1,2\rparen}$$\begin{gathered} A=z_{xx}(1,2)=-8 \\ C=z_{yy}(1,2)=-2 \\ B=z_{xy}(1,2)=0 \end{gathered}$

tHUS;

$\begin{gathered} Since \\ AC-B^2=(-8)(-2)-0^2=16>0 \\ And\text{ A<0} \end{gathered}$

Then (1,2) is a local maximum.

Check for saddle points

$No\text{ saddle points, no local minima}$