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Found in: Page 623

### Essential Calculus: Early Transcendentals

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

# 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 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 $$\left\{ {(x,y,z)/x + y + z \le 10} \right\}$$