Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.
f(x, y, z) = x2 + y2 + z2; x4 + y4 + z4 = 7

Answer 1

Answer:

The maximum value is +√21 and the minimum value is -√21

Step-by-step explanation:

f(x,y,z) = x² + y² + z². Let g(x,y,z) = x⁴ + y⁴ + z⁴ - 7 = 0

Using Lagrange multipliers,

df/dx = λg/dx, df/dy = λg/dy. and df/dz = λg/dz

df/dx = 2x, df/dy = 2y, df/dz = 2z

dg/dx = 4x³, dg/dy = 4y³, dg/dz = 4z³

So, df/dx = λg/dx ⇒ 2x = 4λx³     (1)

df/dy = λg/dy ⇒ 2y = 4λy³           (2)

df/dz = λg/dz ⇒ 2z = 4λz³           (3)

From (1) 4λx³ - 2x = 0

2λx³ - x = 0

x(2λx² - 1) = 0

solving, x = 0 or (2λx² - 1) = 0 ⇒ 2λx² = 1 ⇒ x = ±1/√(2λ) since x ≠ 0

From (2) 4λy³ - 2y = 0

2λy³ - y = 0

y(2λy² - 1) = 0

solving, y = 0 or (2λy² - 1) = 0 ⇒ 2λy² = 1 ⇒ y = ±1/√(2λ) since y ≠ 0

From (3) 4λz³ - 2z = 0

2λz³ - z = 0

z(2λz² - 1) = 0

solving, z = 0 or (2λz² - 1) = 0 ⇒ 2λz² = 1 ⇒ z = ±1/√(2λ) since z ≠ 0

g(x,y,z) = x⁴ + y⁴ + z⁴ - 7 = 0

(1/√(2λ))⁴ + (1/√(2λ))⁴ + (1/√(2λ))⁴ - 7 = 0

3 (1/√(2λ))⁴ = 7

(1/√(2λ))⁴ = 7/3

1/√(2λ) = ⁴√7/3

√(2λ) = ⁴√3/7

2λ = √3/7

λ = 1/2(√3/7)

Since x = 1/√(2λ) = 1/√(2 [1/2(√3/7)]) = 1/⁴√3/7 = ±⁴√7/3

Also y = 1/√(2λ) = 1/√(2 [1/2(√3/7)]) = 1/⁴√3/7 = ±⁴√7/3

and z = 1/√(2λ) = 1/√(2 [1/2(√3/7)]) = 1/⁴√3/7 = ±⁴√7/3

Substituting x,y and z into f(x,y,z) we have

f(x,y,z) = (⁴√7/3)² + (⁴√7/3)² + (⁴√7/3)² = 3(⁴√7/3)² = 3(√7/3) = √(7 × 3) = ±√21

The maximum value is +√21 and the minimum value is -√21