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
