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: 22

Answer

There is no critical point for $f$. Hence, there is no local maximum, no local minimum nor saddle point.

Work Step by Step

We have $f\left( {x,y} \right) = \left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$. The partial derivatives are ${f_x} = {{\rm{e}}^{{x^2} - {y^2}}} + 2x\left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$ $ = \left( {1 + 2{x^2} - 2xy} \right){{\rm{e}}^{{x^2} - {y^2}}}$ ${f_y} = - {{\rm{e}}^{{x^2} - {y^2}}} - 2y\left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$ $ = - \left( {1 + 2xy - 2{y^2}} \right){{\rm{e}}^{{x^2} - {y^2}}}$ ${f_{xx}} = 2x{{\rm{e}}^{{x^2} - {y^2}}} + \left( {4x - 2y} \right){{\rm{e}}^{{x^2} - {y^2}}} + 4{x^2}\left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$ $ = 2{{\rm{e}}^{{x^2} - {y^2}}}\left( {3x + 2{x^3} - y - 2{x^2}y} \right)$ ${f_{yy}} = 2y{{\rm{e}}^{{x^2} - {y^2}}} + \left( { - 2x + 4y} \right){{\rm{e}}^{{x^2} - {y^2}}} + 4{y^2}\left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$ $ = 2{{\rm{e}}^{{x^2} - {y^2}}}\left( {3y - 2{y^3} + - x + 2x{y^2}} \right)$ ${f_{xy}} = - 2y{{\rm{e}}^{{x^2} - {y^2}}} - 2x{{\rm{e}}^{{x^2} - {y^2}}} - 4xy\left( {x - y} \right){{\rm{e}}^{{x^2} - {y^2}}}$ $ = - 2{{\rm{e}}^{{x^2} - {y^2}}}\left( {x + y + 2{x^2}y - 2x{y^2}} \right)$ To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$: ${f_x} = \left( {1 + 2{x^2} - 2xy} \right){{\rm{e}}^{{x^2} - {y^2}}} = 0$ ${f_y} = - \left( {1 + 2xy - 2{y^2}} \right){{\rm{e}}^{{x^2} - {y^2}}} = 0$ Since ${{\rm{e}}^{{x^2} - {y^2}}} \ne 0$, we obtain two simultaneous equations (1) ${\ \ \ }$ $1 + 2{x^2} - 2xy = 0$ ${\ }$ and ${\ }$ $ - 1 - 2xy + 2{y^2} = 0$ Adding the two equations of equation (1) gives $2{x^2} + 2{y^2} - 4xy = 0$ ${x^2} + {y^2} - 2xy = 0$ ${\left( {x - y} \right)^2} = 0$ The solution is $y=x$. Subtracting the two equations of equation (1) gives $2{x^2} - 2{y^2} + 2 = 0$ ${x^2} - {y^2} + 1 = 0$ Substituting $y=x$ in the last equation gives $1=0$. Thus, there is no solution. Therefore, we conclude that there is no critical point for $f$. Hence, there is no local maximum, no local minimum nor saddle point.
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