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Q4E

Expert-verifiedFound in: Page 623

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.**

- The g(1,2,3)=24
- The domain of g is bounded below the plane x+y+z=10

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

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

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

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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