# Find and classify all the critical points of the function g(x,y)=Ax^3+By^3+Cxy. Make up your own...

## Question:

Find and classify all the critical points of the function {eq}g(x,y)=Ax^3+By^3+Cxy{/eq}. Make up your own function in the same vein as this one, then find and cassify its crtitical points. More interesting functions will be harder, but also worth more points.

## Second Derivative Test:

Let {eq}\left( a,b \right) {/eq} be a critical point for a two-variable function {eq}f\left( x,y \right) {/eq}. We define a discriminant {eq}D\left( x,y \right)={{f}_{xx}}{{f}_{yy}}-{{\left[ {{f}_{xy}} \right]}^{2}} {/eq}. If {eq}D>0,{{f}_{xx}}>0 {/eq}, then {eq}\left( a,b \right) {/eq} is a relative minimum. If {eq}D>0,{{f}_{xx}}<0 {/eq}, then {eq}\left( a,b \right) {/eq} is a relative maximum. If {eq}D<0,\left( a,b \right) {/eq} is a saddle point (also called a minimax point). If {eq}D=0 {/eq}, the second derivative test fails to give any information about critical point.

The given function is {eq}g\left( x,y \right)=A{{x}^{3}}+B{{y}^{3}}+Cxy {/eq}.

We will need the following.

{eq}\begin{align} & {{g}_{x}}=3A{{x}^{2}}+Cy=0.....\left( 1 \right) \\ & {{g}_{y}}=3B{{y}^{2}}+Cx=0.....\left( 2 \right) \\ \end{align} {/eq}

To get the critical point we solve {eq}\left( 1 \right) {/eq} & {eq}\left( 2 \right) {/eq}.

From {eq}\left( 2 \right),x=\frac{-3B{{y}^{2}}}{C} {/eq}

Put {eq}x {/eq} in {eq}\left( 1 \right) {/eq}

{eq}\begin{align} & 3A{{\left( \frac{-3B{{y}^{2}}}{C} \right)}^{2}}+Cy=0 \\ & \Rightarrow \frac{27A{{B}^{2}}}{{{C}^{2}}}{{y}^{4}}+Cy=0 \\ & \Rightarrow y\left[ \frac{27A{{B}^{2}}}{{{c}^{2}}}{{y}^{3}}+C \right]=0 \\ & \Rightarrow y=0\text{ or }\frac{27A{{B}^{2}}}{{{C}^{2}}}{{y}^{3}}+C=0 \\ & \Rightarrow {{y}^{3}}=\frac{-{{C}^{3}}}{27A{{B}^{2}}} \\ & y={{\left( \frac{-{{C}^{3}}}{27A{{B}^{2}}} \right)}^{{}^{1}\!\!\diagup\!\!{}_{3}\;}} \\ \end{align} {/eq}

We will require numerical values of {eq}A,B {/eq} and {eq}C {/eq} to get the critical points and classify them.

Let {eq}A=1,B=\sqrt{8},C=-3 {/eq}.

{eq}\begin{align} & y={{\left( \frac{-{{\left( -3 \right)}^{3}}}{27\left( 1 \right){{\left( \sqrt{8} \right)}^{2}}} \right)}^{{}^{1}\!\!\diagup\!\!{}_{3}\;}} \\ & ={{\left( \frac{1}{8} \right)}^{{}^{1}\!\!\diagup\!\!{}_{3}\;}} \\ & =\frac{1}{2},-\frac{1}{2},-\frac{1}{2} \\ \end{align} {/eq}

Thus, {eq}x=\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} {/eq}.

Thus, the critical points are {eq}\left( \frac{1}{\sqrt{2}},\frac{1}{2} \right),\left( \frac{1}{\sqrt{2}},-\frac{1}{2} \right) {/eq} and {eq}\left( \frac{1}{\sqrt{2}},-\frac{1}{2} \right) {/eq}

We classify the critical points.

The function equation becomes.

{eq}\begin{align} & g\left( x,y \right)={{x}^{3}}+\sqrt{8}{{y}^{3}}-3xy \\ & {{g}_{x}}=3{{x}^{2}}-3y \\ & {{g}_{xx}}=6x \\ & {{g}_{y}}=3\sqrt{8}{{y}^{2}}-3x \\ & {{g}_{yy}}=6\sqrt{8}y \\ & {{g}_{xy}}=-3 \\ \end{align} {/eq}

The discriminant is

{eq}\begin{align} & D={{g}_{xx}}{{g}_{yy}}-{{\left[ {{g}_{xy}} \right]}^{2}} \\ & =\left( 6x \right)\left( 3\sqrt{8}{{y}^{2}}-3x \right)-{{\left( -3 \right)}^{2}} \\ & =18\sqrt{8}x{{y}^{2}}-18{{x}^{2}}-9 \\ \end{align} {/eq}

At {eq}\left( \frac{1}{\sqrt{2}},\frac{1}{2} \right) {/eq},

{eq}D=-9<0 {/eq}

Hence, {eq}\left( \frac{1}{\sqrt{2}},\frac{1}{2} \right) {/eq} is a saddle point.

At {eq}\left( \frac{1}{\sqrt{2}},-\frac{1}{2} \right) {/eq}

{eq}D=-9<0 {/eq}

Hence, this point is also a saddle point.

The third critical point is a repeated root. Hence, all the critical points of {eq}g {/eq} are saddle points.

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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