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Essential Calculus: Early Transcendentals
Found in: Page 623
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Let \(g(x,y,z) = {x^3}{y^2}z\sqrt {10 - x - y - z} \)

a) Evaluate g(1,2,3)

b) Find and describe the domain of g.

  1. The g(1,2,3)=24
  2. The domain of g is bounded below the plane x+y+z=10
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1:- Consideration

Let \(g(x,y,z) = {x^3}{y^2}z\sqrt {10 - x - y - z} \) is a function

Step 2:- Substitution

Let x=1,y=2,z=3 to find g(1,2,3)

b)The domain of the g is bounded below the plane x+y+z=10

Step 1:- Finding the relation:

For the function g to exist

\({x^3}{y^2}z\sqrt {10 - x - y - z} \ge 0\)

Which means x,y,z must belong to real numbers and

\(\begin{aligned}{l}10 - x - y - z \ge 0\\ &\Rightarrow - (x + y + z) \ge - 10\\ &\Rightarrow x + y + z \le 10\end{aligned}\)

Therefore, the domain of \({x^3}{y^2}z\sqrt {10 - x - y - z} \) is \(\left\{ {(x,y,z)/x + y + z \le 10} \right\}\)

Step 2: Description

The region below the plane x+y+z=10 is represent as \(x + y + z < 10\) and bounded with the plane:

Therefore, the domain of \({x^3}{y^2}z\sqrt {10 - x - y - z} \) is \(\left\{ {(x,y,z)/x + y + z \le 10} \right\}\)

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