Question

find the minimum value of x^2+y^2+z^2subject to the condition 1/x+1/y+1/z=1​

Answer 1

Answer:

Step-by-step explanation:

1)2x=λ(1)+μ(yz)

2)2y=λ(1)+μ(xz)

3)2z=λ(1)+μ(xy)

Multiplying (1) by x, (2) by y and , (3) by z I get:

2(x2+y2+z2)=λ(1)+μ(3xyz)

[Using 3xyz=−3]

2u=λ−3μ where u is the function to be maximized/minimized. λ and μ are constants.

After this I'm not sure how to proceed to find maximum/minimum value of u. Any suggestions?

Answer 2

Answer: The minimum value of  $x^2+y^2+z^2$ the subject to the condition $1/x+1/y+1/z=1$ is  $\frac{1}{2(xy+yz+xz)}(x^2+y^2+z^2)=\frac{1}{2}$.

The minimum value of $x^2+y^2+z^2$ the subject to the condition $1/x+1/y+1/z=1$ can be found by using Lagrange multipliers.

Let $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)=1/x+1/y+1/z$.

Then the minimum value of $f(x,y,z)$ subject to the condition $g(x,y,z)=1$ can be found by solving the system of equations

            $2x\lambda1=\lambda2 \ 2y\lambda1=\lambda3 \ 2z\lambda1=\lambda4 \ \lambda2x+\lambda3y+\lambda_4z=1 \end{cases}$

Solving this system of equations gives us

$\lambda1=\frac{1}{2(xy+yz+xz)}, \quad \lambda2=\frac{xy}{xy+yz+xz}, \quad \lambda3=\frac{yz}{xy+yz+xz}, \quad \lambda4=\frac{xz}{xy+yz+xz}$

Therefore, The minimum value of  $x^2+y^2+z^2$ the subject to the condition $1/x+1/y+1/z=1$ is  $\frac{1}{2(xy+yz+xz)}(x^2+y^2+z^2)=\frac{1}{2}$.

Learn more about Lagrange multipliers here

https://brainly.in/question/37367872

Learn more about  minimum value here

https://brainly.in/question/48266302

#SPJ2