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allochka39001 [22]
1 year ago
12

guse lagrange multipliers to find the maximum or minimum values of the function subject to the given constraint. (if an answer d

oes not exist, enter dne.) f(x, y)
Mathematics
1 answer:
Novosadov [1.4K]1 year ago
3 0

Therefore the maximum value of function f(x,y,z)=x^{2} y^{2} z^{2} =1/27

And the minimum value is 0

<h3>What is function?</h3>

function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable) (the dependent variable) (the dependent variable) (the dependent variable). Mathematics uses functions frequently, and functions are essential for specifying physical relationships in the sciences.

Here,

The function is given as:

f(x,y,z)=x^{2} y^{2} z^{2}

x^{2} +y^{2}+ z^{2}=1

=>x^{2} +y^{2}+ z^{2}-1=0

Using Lagrange multiplies, we have:

L(x,y,z,λ)=f(x,y,z) +λ(0)

Substitute f(x,y,z)=x^{2} y^{2} z^{2}  and x^{2} +y^{2}+ z^{2}-1=0

Differentiate

L(x)=2xy^{2} z^{2}+2λx

L(y)=2yx^{2} z^{2}+2λy

L(z)=2zx^{2} y^{2}+2λz

L(λ)=x^{2} +y^{2}+ z^{2}-1

Equating to 0

2xy^{2} z^{2}+2λx =0

2yx^{2} z^{2}+2λy = 0

2zx^{2} y^{2}+2λz = 0

x^{2} +y^{2}+ z^{2}-1 = 0

Factorize the above expressions

2xy^{2} z^{2}+2λx =0

2x(y^{2} z^{2}+λ)=0

2x=0 and (y^{2} z^{2}+λ)=0

x=0 and  y^{2} z^{2}= -λ

2yx^{2} z^{2}+2λy = 0

2y(x^{2} z^{2}+λ)=0

2y=0 and (x^{2} z^{2}+λ)=0

y=0 and  x^{2} z^{2}= -λ

2zx^{2} y^{2}+2λz = 0

2z(y^{2} x^{2}+λ)=0

2z=0 and (x^{2} y^{2}+λ)=0

z=0 and  x^{2} y^{2}= -λ

So we have ,

x=0 and  y^{2} z^{2}= -λ

y=0 and  x^{2} z^{2}= -λ

z=0 and  x^{2} y^{2}= -λ

The above expression becomes

x=y=z=0

This means that,

x^{2} +y^{2}+ z^{2}=1

x^{2} +x^{2}+ x^{2} =1 \\3x^{2 } =1

x= ±1/\sqrt{3}

So,

y= ±1/\sqrt{3}

z= ±1/\sqrt{3}

The critic points are

x=y=z=±1/\sqrt{3}

x=y=z=0

Therefore the maximum value of function f(x,y,z)=x^{2} y^{2} z^{2} =1/27

And the minimum value is 0

To know  more about function , visit

brainly.com/question/12426369

#SPJ4

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Answer:

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