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 - 13.10 Exercises - Page 958: 7

Answer

$f\left( {1,1} \right) = 2$

Work Step by Step

$$\eqalign{ & {\text{Maximize }}f\left( {x,y} \right) = \sqrt {6 - {x^2} - {y^2}} \cr & {\text{Constraint: }}x + y - 2 = 0{\text{ }} \to {\text{ }}g\left( x \right) = x + y - 2 \cr & {\text{Let }}f\left( {x,y} \right) = \sqrt {h\left( x \right)} ,{\text{ }}h\left( x \right) = 6 - {x^2} - {y^2} \cr & {\text{then }}f\left( {x,y} \right){\text{ is maximum when }}h\left( x \right){\text{ is maximum}}{\text{.}} \cr & \cr & {\text{We need to maximize }}h\left( x \right) = 6 - {x^2} - {y^2}{\text{ }}\left( {{\text{objective function}}} \right) \cr & {\text{Using the Method of Lagrange Multipliers}} \cr & *{\text{ }}{h_x}\left( {x,y} \right) = \lambda {g_x}\left( {x,y} \right) \cr & \frac{\partial }{{\partial x}}\left[ {6 - {x^2} - {y^2}} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {x + y - 2} \right] \cr & - 2x = \lambda {\text{ }}\left( {\bf{1}} \right) \cr & \cr & *{\text{ }}{h_y}\left( {x,y} \right) = \lambda {g_y}\left( {x,y} \right) \cr & \frac{\partial }{{\partial y}}\left[ {6 - {x^2} - {y^2}} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {x + y - 2} \right] \cr & - 2y = \lambda {\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{ and }}\left( {\bf{2}} \right) \cr & x = y \cr & \cr & *{\text{ }}g\left( {x,y} \right) = c \cr & x + y - 2 = 0 \cr & x + y = 2 \cr & {\text{Where }}x = y \cr & x + x = 2 \cr & x = 1 \to y = 1 \cr & {\text{Therefore}}{\text{,}} \cr & f\left( {1,1} \right) = \sqrt {6 - {{\left( 1 \right)}^2} - {{\left( 1 \right)}^2}} \cr & f\left( {1,1} \right) = 2 \cr & \cr & {\text{The maximum is }}2{\text{ at }}x = 1,\,y = 1 \cr} $$
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