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: 33

Answer

There is only one critical point at $\left( {0,0} \right)$. However, by the Second Derivative Test, the critical point $\left( {0,0} \right)$ is a saddle point. Therefore, $f$ does not have a global minimum or a global maximum on the domain $D$. This does not contradict Theorem 3 because the domain $D$ is not closed.

Work Step by Step

We have $f\left( {x,y} \right) = xy$, and the domain $D = \left\{ {\left( {x,y} \right):0 < x < 1,{\ \ }0 < y < 1} \right\}$. Notice that it is a open square. Thus, it is bounded but not closed. The partial derivatives are ${f_x} = y$, ${\ \ \ \ }$ ${f_y} = x$ ${f_{xx}} = 0$, ${\ \ }$ ${f_{yy}} = 0$, ${\ \ }$ ${f_{xy}} = 1$ To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$: ${f_x} = y = 0$, ${\ \ \ }$ ${f_y} = x = 0$ So, there is only one critical point at $\left( {0,0} \right)$. We evaluate the discriminant ${f_{xx}}{f_{yy}} - {f_{xy}}^2 = - 1$. By the Second Derivative Test, the critical point $\left( {0,0} \right)$ is a saddle point. However, the critical point $\left( {0,0} \right)$ is not defined in the domain $D = \left\{ {\left( {x,y} \right):0 < x < 1,{\ \ }0 < y < 1} \right\}$. Therefore, we conclude that $f$ does not have a global minimum or a global maximum on $D$. This does not contradict Theorem 3 because the domain $D$ is not closed.
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