Finding and classifying critical points?

[allhalf425]

How would I find and classify the critical points of a function as follows...

Find and classify the critical points of

z = f(x, y) = x[sup:3alyxftz]2[/sup:3alyxftz] - 2xy + y[sup:3alyxftz]3[/sup:3alyxftz] - y

 

[Deleted member 4993]

How would I find and classify the critical points of a function as follows...

Find and classify the critical points of

z = f(x, y) = x[sup:4n7k91er]2[/sup:4n7k91er] - 2xy + y[sup:4n7k91er]3[/sup:4n7k91er] - y

First, you need to know the definition of a critical point of a function.

Please share your work with us, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[soroban]

Hello, allhalf425!

$\displaystyle \text{Find and classify the critical points of }\;f(x,y) \;=\;x^2 - 2xy + y^3 - y$

Set the two partial derivatives equal to zero and solve.

. . $\displaystyle f_x \;=\;2x-2y \:=\:0 \quad\Rightarrow\quad y \:=\:x\;\;[1]$

. . $\displaystyle f_y \:=\:-2x + 3y^2 - 1 \:=\:0\;\;[2]$


$\displaystyle \text{Substitute [1] into [2]: }\;2x + 3x^2 - 1 \:=\:0 \quad\Rightarrow\quad 3x^2 - 2x - 1 \:=\:0$

\(\displaystyle {\text{Factor: }\;(x-1)(3x+1) \:=\:0\)

$\displaystyle \text{And we have: }\;\begin{Bmatrix}x-1 \:=\:0 & \Rightarrow & x \:=\:1 \\ 3x+1\:=\:0 & \Rightarrow & x \:=\:\text{-}\frac{1}{3} \end{Bmatrix}$

$\displaystyle \text{Then: }\;\begin{Bmatrix}y \:=\:1 \\ y \:=\:\text{-}\tfrac{1}{3} \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}f \:=\:\text{-}1 \\ f \:=\:\frac{5}{27} \end{Bmatrix}$



Second Partials Test

$\displaystyle f_{xx} \:=\:2,\quad f_{yy} \:=\:6y,\quad f_{xy} \:=\:-2$

$\displaystyle \text{Then: }\;D \;=\;(f_{xx})(f_{yy}) - (f_{xy})^2 \;=\;(2)(6y) - (-2)^2 \;=\;12y - 4 \;=\;4(y-3)$


\(\displaystyle \text{At }(1,1)\!:\;\;D \:=\:4(1-3) \:=\:-8 \quad\hdots \text{ negative: saddle point at }(1,1,\text{-}1)\)

\(\displaystyle \text{At }\left(\text{-}\tfrac{1}{3},\text{-}\tfrac{1}{3}\right)\!:\;\;D \;=\;4\left(\text{-}\tfrac{1}{3} - 3\ritght) \:=\:\text{-}\tfrac{40}{3}\quad\hdots\text{ negative: saddle point at }\left(-\tfrac{1}{3},\text{-}\tfrac{1}{3},\tfrac{5}{27}\right)\)

 

[BigGlenntheHeavy]

What its graph looks like.

[attachment=0:1wy51hgr]zzz.gif[/attachment:1wy51hgr]