Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 951: 25


$h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$; set of points: {$(x,y) | 2x+3y\geq 6$}

Work Step by Step

As per the given values; $h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$ The term $(2x+3y-6)^2$ is a polynomial and will be always continuous and the term $ \sqrt{ 2x+3y-6}$ is a squared root which only defines for the non-negative values. Thus, $(2x+3y-6)\geq 0$ or, $(2x+3y)\geq 6$ or, $3y\geq -2x+6$ or, $ y\geq \dfrac{-2}{3}x+2$ Hence, $h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$; set of points: {$(x,y) | 2x+3y\geq 6$}
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