Use calculus to find the precise values of the local maximum and minimum and saddle points of the...

Question:

Use calculus to find the precise values of the local maximum and minimum and saddle points of the function. {eq}f(x, y) = 7xye^{-x^2 - y^2}. {/eq}

Method to find Relative maxima and Minima of a function:


Consider a continuous and differentiable function {eq}f(x,y). {/eq} First find the critical points of the given function. Then find the discriminant {eq}D. {/eq} Then

  • Critical Point {eq}(a,b) {/eq} has a relative minima if {eq}D = rt - s^2 > 0 \ \ \ \text{and} \ \ \ r > 0 {/eq}
  • Critical Point {eq}(a,b) {/eq} has a relative maxima if {eq}D = rt - s^2 > 0 \ \ \ \text{and} \ \ \ r < 0 {/eq}
  • Critical Point {eq}(a,b) {/eq} is a saddle point if {eq}D = rt - s^2 < 0 {/eq}

where {eq}r = f_{xx}(a,b) , \ \ s = f_{xy}(a,b) \ \ \text{and} \ \ t = f_{yy}(a,b). {/eq}


Answer and Explanation:


{eq}\hspace{30mm} \displaystyle{ f(x,y) = 7xy e^{-x^2 - y^2} \\ f_{x} = 7y e^{-x^2 - y^2} - 14 x^2 y e^{-x^2 - y^2} \\ f_{y} = 7x e^{-x^2 - y^2} - 14 x y^2 e^{-x^2 - y^2} } {/eq}

Now find the critical points:

{eq}\hspace{30mm} \displaystyle{ f_{x} = 0 \implies 7y(1-2x^2) = 0 \\ f_{y} = 0 \implies 7x (1-2y^2) = 0 } {/eq}

The solution of the above system is the critical points. Therefore the critical points are:

{eq}\hspace{30mm} \displaystyle{ (0,0) , \ \ \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) , \ \ \left( \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right) , \ \ \left( - \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) \ \ \text{and} \ \ \left( - \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right) } {/eq}

{eq}\hspace{30mm} \displaystyle{ r = f_{xx} = 14xy e^{-x^2-y^2} (2x^2 - 3) \\ t = f_{yy} = 14 xy e^{-x^2 - y^2} (2y^2 - 3) \\ s = 7 e^{-x^2-y^2} (1 - 2xy - 2x^2 + 4x^2 y^2) } {/eq}

Now at point {eq}(0,0), {/eq}

{eq}\hspace{30mm} \displaystyle{ r = 0 \\ t = 0 \\ s = 7 \\ D(0,0) = - 49 < 0 } {/eq}

Therefore, point (0,0) is saddle point.


Now at point {eq}\left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right), {/eq}

{eq}\hspace{30mm} \displaystyle{ r = -\frac{14}{e} \\ t = -\frac{14}{e} \\ s = 0 \\ D \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) > 0 \ \ \text{and} \ \ r < 0 } {/eq}

Therefore, point {eq}\left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) {/eq} is local maxima.


Now at point {eq}\left( \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right), {/eq}

{eq}\hspace{30mm} \displaystyle{ r = \frac{14}{e} \\ t = \frac{14}{e} \\ s = \frac{14}{e} \\ D \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) > 0 \ \ \text{and} \ \ r > 0 } {/eq}

Therefore, point {eq}\left( \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right) {/eq} is local minima.


Now at point {eq}\left( - \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right), {/eq}

{eq}\hspace{30mm} \displaystyle{ r = \frac{14}{e} \\ t = \frac{14}{e} \\ s = \frac{14}{e} \\ D \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) > 0 \ \ \text{and} \ \ r > 0 } {/eq}

Therefore, point {eq}\left( - \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) {/eq} is local minima.


Now at point {eq}\left( - \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right), {/eq}

{eq}\hspace{30mm} \displaystyle{ r = - \frac{14}{e} \\ t = - \frac{14}{e} \\ s = 0 \\ D \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) > 0 \ \ \text{and} \ \ r < 0 } {/eq}

Therefore, point {eq}\left( - \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} \right) {/eq} is local maxima.



Learn more about this topic:

Loading...
Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2
20K

Related to this Question

Explore our homework questions and answers library

Support