Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.7 Optimization in Several Variables - Exercises - Page 822: 13

Answer

The critical points and their nature: $\begin{array}{*{20}{c}} {{\rm{Critical}}}&{}\\ {{\rm{Point}}}&{{\rm{Type}}}\\ {\left( {0,0} \right)}&{{\rm{saddle{\ } point}}}\\ {\left( {1,1} \right)}&{{\rm{local{\ } minimum}}}\\ {\left( { - 1, - 1} \right)}&{{\rm{local{\ } minimum}}} \end{array}$

Work Step by Step

We have $f\left( {x,y} \right) = {x^4} + {y^4} - 4xy$. The partial derivatives are ${f_x} = 4{x^3} - 4y$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4x$ ${f_{xx}} = 12{x^2}$, ${\ \ }$ ${f_{yy}} = 12{y^2}$, ${\ \ }$ ${f_{xy}} = - 4$ To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$: ${f_x} = 4{x^3} - 4y = 0$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4x = 0$ From the first equation we obtain $y = {x^3}$. Substituting it in the second equation gives $4{x^9} - 4x = 0$ $4x\left( {{x^8} - 1} \right) = 0$ The solutions are $x=0$, $x = \pm 1$. So, the critical points are $\left( {0,0} \right)$, $\left( {1,1} \right)$, $\left( { - 1, - 1} \right)$. Next, we use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table: $\begin{array}{*{20}{c}} {{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\rm{Discriminant}}}&{}\\ {{\rm{Point}}}&{12{x^2}}&{12{y^2}}&{ - 4}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\ {\left( {0,0} \right)}&0&0&{ - 4}&{ - 16}&{{\rm{saddle{\ } point}}}\\ {\left( {1,1} \right)}&{12}&{12}&{ - 4}&{128}&{{\rm{local{\ } minimum}}}\\ {\left( { - 1, - 1} \right)}&{12}&{12}&{ - 4}&{128}&{{\rm{local{\ } minimum}}} \end{array}$
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