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 852: 3



Work Step by Step

Given: $f(x,y) =49-x^2-y^2$ $g(x,y)=x+3y-10=0$ ...(1) Consider the gradient equation as follows: $\nabla =\lambda \nabla g$ Here, $ -2x=\lambda$ ; $ -2y=3 \lambda$ This implies that $x=-\dfrac{ \lambda}{2}; y=-\dfrac{3\lambda}{2}$ Plugging $x=-\dfrac{ \lambda}{2}; y=-\dfrac{3\lambda}{2}$ in equation (1), we get $g(x,y)=x+3y-10=0$ This implies that $\lambda=-2$ and $x=1 ; y=3$ Hence, $f(x,y)=49-x^2-y^2 =49-1-9=39$
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