Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 13

Answer

$${\text{Relative minimum at the point }}\left( {3, - 4, - 5} \right)$$

Work Step by Step

$$\eqalign{ & z = {x^2} + xy + \frac{1}{2}{y^2} - 2x + y \cr & {\text{Let }}z = f\left( {x,y} \right) \cr & f\left( {x,y} \right) = {x^2} + xy + \frac{1}{2}{y^2} - 2x + y \cr & {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + xy + \frac{1}{2}{y^2} - 2x + y} \right] \cr & {f_x}\left( {x,y} \right) = 2x + y - 2 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + xy + \frac{1}{2}{y^2} - 2x + y} \right] \cr & {f_y}\left( {x,y} \right) = x + y + 1 \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr & 2x + y - 2 = 0,{\text{ }}x + y + 1 = 0 \cr & {\text{Solving the system of equations, we obtain}} \cr & x = 3,{\text{ }}y = - 4 \cr & {\text{The critical point is }}\left( {3, - 4} \right) \cr & {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x + y - 2} \right] = 2 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x + y + 1} \right] = 1 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + y - 2} \right] = 1 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {3, - 4} \right) \cr & d\left( {3, - 4} \right) = {f_{xx}}\left( {3, - 4} \right){f_{yy}}\left( {3, - 4} \right) - {\left[ {{f_{xy}}\left( {3, - 4} \right)} \right]^2} \cr & d\left( {3, - 4} \right) = \left( 2 \right)\left( 1 \right) - {\left[ 1 \right]^2} \cr & d\left( {3, - 4} \right) = 1 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {3, - 4} \right) = 1 > 0 \cr & f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {3, - 4,f\left( {3, - 4} \right)} \right) \cr & f\left( {x,y} \right) = {x^2} + xy + \frac{1}{2}{y^2} - 2x + y \cr & f\left( {3, - 4} \right) = {\left( 3 \right)^2} + \left( 3 \right)\left( { - 4} \right) + \frac{1}{2}{\left( { - 4} \right)^2} - 2\left( 3 \right) + \left( { - 4} \right) \cr & f\left( {3, - 4} \right) = - 5 \cr & {\text{Relative minimum at the point }}\left( {3, - 4, - 5} \right) \cr} $$
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