Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.
$f (x, y, z) = x + y + z when x^{2} + 4 y^{2} + 16 z^{2} = 64$

The maximum value of the function is $2 \sqrt{21}$ and the minimum value is $- 2 \sqrt{21}$ and both exist because the constraint is a bounded and closed ellipsoid.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Given function is $f (x, y, z) = x + y + z .$

Given constraint is $x^{2} + 4 y^{2} + 16 z^{2} = 64 .$

Step 2. critical points of the function.

Gradients of function.

$\nabla f (x, y, z) = i + j + k \nabla g (x, y, z) = 2 x i + 8 y j + 32 z k$

Use the method of Lagrange multipliers.

$\nabla f (x, y, z) = λ \nabla g (x, y, z) i + j + k = λ (2 x i + 8 y j + 32 z k) i + j + k = 2 λ x i + 8 λ y j + 32 λ z k$

Compare terms.

$1 = 2 λ x \Rightarrow λ = \frac{1}{2 x} 1 = 8 λ y \Rightarrow λ = \frac{1}{8 y} 1 = 32 λ z \Rightarrow λ = \frac{1}{32 z} so x = 4 y & z = \frac{y}{4}$

substitute $x = 4 y & z = \frac{y}{4}$ in constraint.

$x^{2} + 4 y^{2} + 16 z^{2} = 64 {(4 y)}^{2} + 4 y^{2} + 16 {(\frac{y}{4})}^{2} = 64 21 y^{2} = 64 y = \pm \frac{8}{\sqrt{21}} x = \pm \frac{32}{\sqrt{21}} z = \pm \frac{2}{\sqrt{21}}$

so critical points are $(\frac{- 32}{\sqrt{21}}, \frac{- 8}{\sqrt{21}}, \frac{- 2}{\sqrt{21}}) & (\frac{32}{\sqrt{21}}, \frac{8}{\sqrt{21}}, \frac{2}{\sqrt{21}}) .$

Step 3. maximum and minimum of a function.

Find function value at $(\frac{- 32}{\sqrt{21}}, \frac{- 8}{\sqrt{21}}, \frac{- 2}{\sqrt{21}}) & (\frac{32}{\sqrt{21}}, \frac{8}{\sqrt{21}}, \frac{2}{\sqrt{21}}) .$

role="math" localid="1649896217514" $f (x, y, z) = x + y + z f (\frac{- 32}{\sqrt{21}}, \frac{- 8}{\sqrt{21}}, \frac{- 2}{\sqrt{21}}) = (\frac{- 32}{\sqrt{21}} + \frac{- 8}{\sqrt{21}} + \frac{- 2}{\sqrt{21}}) f (\frac{- 32}{\sqrt{21}}, \frac{- 8}{\sqrt{21}}, \frac{- 2}{\sqrt{21}}) = \frac{- 42}{\sqrt{21}} f (\frac{- 32}{\sqrt{21}}, \frac{- 8}{\sqrt{21}}, \frac{- 2}{\sqrt{21}}) = - 2 \sqrt{21} and f (\frac{32}{\sqrt{21}}, \frac{8}{\sqrt{21}}, \frac{2}{\sqrt{21}}) = 2 \sqrt{21}$

So the maximum value of the function is $2 \sqrt{21}$ and the minimum value is $- 2 \sqrt{21} .$

As constraint is bounded and closed ellipsoid so maximum and minimum both exist.

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Calculus

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Short Answer

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.
$f (x, y, z) = x + y + z when x^{2} + 4 y^{2} + 16 z^{2} = 64$

Step by Step Solution

Step 1. Given information.

Step 2. critical points of the function.

Step 3. maximum and minimum of a function.

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Short Answer

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.f(x,y,z)=x+y+z when x2+4y2+16z2=64

Step by Step Solution

Step 1. Given information.

Step 2. critical points of the function.

Step 3. maximum and minimum of a function.

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In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.
$f (x, y, z) = x + y + z when x^{2} + 4 y^{2} + 16 z^{2} = 64$