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: 6

Answer

$f\left( {\root 3 \of 4 ,\frac{1}{{\root 3 \of 2 }}} \right) = 3\root 3 \of 4 + \frac{1}{{\root 3 \of 2 }} + 10$

Work Step by Step

$$\eqalign{ & {\text{Minimize }}f\left( {x,y} \right) = 3x + y + 10 \cr & {\text{Constraint: }}{x^2}y = 6{\text{ }} \to {\text{ }}g\left( x \right) = {x^2}y - 6 \cr & \cr & {\text{We need to minimize }}f\left( x \right) = 3x + y + 10{\text{ }}\left( {{\text{objective function}}} \right) \cr & {\text{Using the Method of Lagrange Multipliers}} \cr & *{\text{ }}{f_x}\left( {x,y} \right) = \lambda {g_x}\left( {x,y} \right) \cr & \frac{\partial }{{\partial x}}\left[ {3x + y + 10} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {{x^2}y - 6} \right] \cr & 3 = \lambda \left( {2xy} \right) \cr & 3 = 2\lambda xy{\text{ }}\left( {\bf{1}} \right) \cr & \cr & *{\text{ }}{f_y}\left( {x,y} \right) = \lambda {g_y}\left( {x,y} \right) \cr & \frac{\partial }{{\partial y}}\left[ {3x + y + 10} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {{x^2}y - 6} \right] \cr & 1 = \lambda \left( {{x^2}} \right) \cr & 1 = \lambda {x^2}{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{ and }}\left( {\bf{2}} \right) \cr & {\text{From the equation }}\left( {\bf{1}} \right),{\text{ }}\lambda = \frac{3}{{2xy}} \cr & {\text{From the equation }}\left( {\bf{2}} \right),{\text{ }}\lambda = \frac{1}{{{x^2}}} \cr & {\text{Equating and solving for }}y \cr & \frac{3}{{2xy}} = \frac{1}{{{x^2}}} \cr & \frac{{2xy}}{3} = {x^2} \cr & y = \frac{{3x}}{2} \cr & {\text{Using the constraint }}{x^2}y = 6 \cr & \frac{6}{{{x^2}}} = \frac{{3x}}{2} \cr & 3{x^3} = 12 \cr & {x^3} = 4 \cr & x = \root 3 \of 4 \cr & and \cr & y = \frac{{3x}}{2} \to y = \frac{{3\root 3 \of 4 }}{2} \cr & y = \frac{1}{{\root 3 \of 2 }} \cr & \cr & {\text{Therefore}}{\text{,}} \cr & f\left( {\root 3 \of 4 ,\frac{1}{{\root 3 \of 2 }}} \right) = 3\root 3 \of 4 + \frac{1}{{\root 3 \of 2 }} + 10 \cr & \cr & {\text{The minimum is }}3\root 3 \of 4 + \frac{1}{{\root 3 \of 2 }} + 10{\text{ at }}x = \root 3 \of 4 ,\,y = \frac{1}{{\root 3 \of 2 }} \cr} $$
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