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: 19

Answer

The critical point and its nature: $\begin{array}{*{20}{c}} {{\rm{Critical}}}&{}\\ {{\rm{Point}}}&{{\rm{Type}}}\\ {\left( {1,\frac{1}{2}} \right)}&{{\rm{local{\ } maximum}}} \end{array}$

Work Step by Step

We have $f\left( {x,y} \right) = \ln x + 2\ln y - x - 4y$. The partial derivatives are ${f_x} = \frac{1}{x} - 1$, ${\ \ \ }$ ${f_y} = \frac{2}{y} - 4$ ${f_{xx}} = - \frac{1}{{{x^2}}}$, ${\ \ }$ ${f_{yy}} = - \frac{2}{{{y^2}}}$, ${\ \ }$ ${f_{xy}} = 0$ To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$: ${f_x} = \frac{1}{x} - 1 = 0$, ${\ \ }$ ${f_y} = \frac{2}{y} - 4 = 0$ The solutions are $x=1$ and $y = \frac{1}{2}$. So, there is only one critical point: $\left( {1,\frac{1}{2}} \right)$. We use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table: $\begin{array}{*{20}{c}} {{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\rm{Discriminant}}}&{}\\ {{\rm{Point}}}&{ - \frac{1}{{{x^2}}}}&{ - \frac{2}{{{y^2}}}}&0&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\ {\left( {1,\frac{1}{2}} \right)}&{ - 1}&{ - 8}&0&8&{{\rm{local{\ } maximum}}} \end{array}$
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