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}`