Answer :
The minimum and the maximum values are 0 and 1/27, respectively.
The extreme values of a function are the minimum and the maximum values of the function.
The function is given as:
f(x, y, z) = x²y²z²
x² + y² + z² = 1
Subtract 1 from both sides of x² + y² + z² – 1
x² + y² + z²-1=0
Using Lagrange multiplies, we have:
L(xy.z. λ) f(x. y.z) + λ(0)
Substitute f(x.y. z) = x³y²z² and x² + y² + z² − 1 = 0
L(x, y, λ) = x²y²z² + λ(x² + y² + 2² – 1)
Differentiate
Lx = 2xy²z² + 2λx
Ly = 2x²yz² + 2λy
Lz = 2x²y³z + 2λz
Lλ = x² + y² + z²
Equate to 0
2xy²z² + 2λx = 0
2x²yz² + 2λy = 0
2x²y²z+ 2λz = 0
x² + y² + z²-1=0
Factorize the above expressions
2xy²z²+2λx = 0
2x(y²z² + λ) = 0
2x = 0 or y²z²+λ=0
x=0 or y²z² = -λ
2x²yz² + 2λy = 0
2y(x²z² + λ) = 0
2y=0 or x²z²+λ=0
y = 0 or x²z² = -λ
2x²y²z+λz = 0
2z(x²y² + λ) = 0
2z = 0 or x²y² +λ=0
z = 0 or x²y² = -λ
So, we have:
x = 0 or y²z² = -λ
y=0 or x²z² = -λ
z = 0 or x²y² = -λ
Because
-λ= -λ
The above expressions become:
x=y=z=0
x²z² = y²z²
x² = y²
x = ±y
Also, we have:
x²y² = x²z²
y² = z²
y = tz
Also:
y²z² = x²y²
z² = x²
z = ±x
So, we have:
x = ±y
y = ±z
z = ±x
This means that:
x=y=z
Recall that: x² + y² + z² = 1
So, we have:
x² + x² + x² = 1
3x² = 1
Divide through by 3
x² = 1/3
Take square roots of both sides
x=±[tex]\frac{1} \sqrt{3}[/tex]
So, we have:
x=y=z=±[tex]\frac{1} \sqrt{3}[/tex]
So, the critical points are:
x=y=z=0
x=y=z=±[tex]\frac{1} \sqrt{3}[/tex]
Substitute the above values in f(x, y, z)=x²y²z²
f(0,0,0) = 0² × 0² × 0² = 0
f(±[tex]\frac{1} \sqrt{3}[/tex],±[tex]\frac{1} \sqrt{3}[/tex],±[tex]\frac{1} \sqrt{3}[/tex]) = (±±[tex]\frac{1} \sqrt{3}[/tex])² × (±±[tex]\frac{1} \sqrt{3}[/tex])² × (±±[tex]\frac{1} \sqrt{3}[/tex])² =[tex]\frac{1}{27}[/tex]
Considering the above values, we have:
Minimum 0
Maximum = 1/27
Hence, the minimum and the maximum values are 0 and 1/27, respectively.
Read more about extreme values at;
brainly.com/question/1286349
#SPJ4
