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


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.

Answer and Explanation:

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.

Learn more about this topic:

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9

Related to this Question

Explore our homework questions and answers library