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: 8

Answer

$f\left( {\frac{3}{2},3} \right) = \frac{{3\sqrt 5 }}{2}$

Work Step by Step

$$\eqalign{ & {\text{Minimize }}f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & {\text{Constraint: }}2x + 4y - 15 = 0 \cr & \cr & {\text{Let }}f\left( {x,y} \right) = \sqrt {h\left( x \right)} ,{\text{ }}h\left( x \right) = {x^2} + {y^2} \cr & {\text{then }}f\left( {x,y} \right){\text{ is minimum when }}h\left( x \right){\text{ is minimum}}{\text{.}} \cr & \cr & {\text{We need to minimize }}h\left( x \right) = {x^2} + {y^2}{\text{ }}\left( {{\text{objective function}}} \right) \cr & {\text{Using the Method of Lagrange Multipliers}} \cr & *{\text{ }}{h_x}\left( {x,y} \right) = \lambda {g_x}\left( {x,y} \right) \cr & \frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {2x + 4y - 15} \right] \cr & 2x = \lambda \left( 2 \right) \cr & x = \lambda {\text{ }}\left( {\bf{1}} \right) \cr & \cr & *{\text{ }}{h_y}\left( {x,y} \right) = \lambda {g_y}\left( {x,y} \right) \cr & \frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {2x + 4y - 15} \right] \cr & 2y = \lambda \left( 4 \right) \cr & y = 2\lambda {\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{ and }}\left( {\bf{2}} \right) \cr & y = 2x \cr & \cr & *{\text{ }}g\left( {x,y} \right) = c \cr & 2x + 4y = 15 \cr & 2x + 4\left( {2x} \right) = 15 \cr & 10x = 15 \cr & x = \frac{3}{2} \cr & \cr & {\text{Where }}y = 2x \cr & y = 3 \cr & \cr & {\text{Therefore}}{\text{,}} \cr & f\left( {\frac{3}{2},3} \right) = \sqrt {{{\left( {\frac{3}{2}} \right)}^2} + {{\left( 3 \right)}^2}} \cr & f\left( {\frac{3}{2},3} \right) = \frac{{3\sqrt 5 }}{2} \cr & \cr & {\text{The minimum is }}\frac{{3\sqrt 5 }}{2}{\text{ at }}x = \frac{3}{2},\,y = 3 \cr} $$
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