Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.8 - Lagrange Multipliers - Exercises 14.8 - Page 853: 26

Answer

4096/25$\sqrt5$

Work Step by Step

f(x,y,z) = xyz g(x,y,z) = x + y + z^{2} -16 ∇f = yzi + xzj + xyk ∇g = i + j + 2zk ∇f = λ∇g yzi + xzj + xyk = λ(i + j + 2zk) yz = λ ,xz = λ , xy = 2zλ -- (1) either z = 0 or y = x As product can't be 0 z $\ne$ 0 , y = x In (1) x$^{2}$ = 2zλ , xz = λ x$^{2}$ = 2z(xz) x = 0 or x = 2z$^{2}$ x = 2z$^{2}$ , y = 2z$^{2}$ --(2) 2z$^{2}$ + 2z$^{2}$ + z$^{2}$ = 16 z = ± 4/$\sqrt 5$ Using this value in (2) x = 32/5 and y = 32/5 f(32/5 , 32/5 , 4/$\sqrt5$) = 4096/25$\sqrt5$
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