Answered: 4. The function has a maximum and…

4. The function has a maximum and minimum extreme value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x,y) = x² + y² + z², xyz = 4

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter12: Algebra Of Matrices

Section12.CR: Review Problem Set

Problem 35CR: Maximize the function fx,y=7x+5y in the region determined by the constraints of Problem 34.

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Question

4.
The function has a maximum and minimum extreme value. Use Lagrange multipliers to find the extreme values of the function
subject to the given constraint.
f(x,y) = x² + y² + z², xyz = 4

Expert Solution

Step 1

Given $f (x, y, z) = x^{2} + y^{2} + z^{2}$ and constraint $x y z = 4$

Let $g (x, y, z) = x y z - 4$

$\frac{\partial f}{\partial x} = λ \frac{\partial g}{\partial x} \frac{\partial f}{\partial y} = λ \frac{\partial g}{\partial y} \frac{\partial f}{\partial z} = λ \frac{\partial g}{\partial z}$

where $λ$ is the lagrange multiplier.

$2 x = λ y z . . . (1) 2 y = λ x z . . . (2) 2 z = λ x y . . . (3)$

Step 2

Solving by multiplying both sides of the equation (1), (2) and (3) by x, y and z respectively,

$2 x^{2} = λ x y z 2 y^{2} = λ x y z 2 z^{2} = λ x y z$

Equating,

$2 x^{2} = 2 y^{2} = 2 z^{2} x^{2} = y^{2} = z^{2} x = y = z$

Substituting back to the constrained equation,

$(x) (x) (x) = 4 x^{3} = 4 x = \sqrt[3]{4} = 1.58 = y = z$

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