Question

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If a value does not exist, enter NONE.) f(x,y,z) = x2y2z2; x2 + y2 + z2 = 1

Answer 1

The Lagrangian is

$L(x,y,z,\lambda)=x^2y^2z^2+\lambda(x^2+y^2+z^2-1)$

where we assume $\lambda\neq0$, with critical points wherever the partial derivatives are identically zero:

$L_x=2xy^2z^2+2\lambda x=0\implies2x(y^2z^2+\lambda)=0\implies x=0\text{ or }y^2z^2=-\lambda$

(note that the latter case requires $\lambda<0$, and we'll see the same requirement in the next two equations)

$L_y=2x^2yz^2+2\lambda y=0\implies2y(x^2z^2+\lambda)=0\implies y=0\text{ or }x^2z^2=-\lambda$

$L_z=2x^2y^2z+2\lambda z=0\implies2z(x^2y^2+\lambda)=0\implies z=0\text{ or }x^2y^2=-\lambda$

$L_\lambda=x^2+y^2+z^2-1=0$

If either $x=0$, $y=0$, or $z=0$, then we can throw out the latter cases for $L_x=0$, $L_y=0$, and $L_z=0$, since we don't want $\lambda=0$. If, for instance, we pick $x=y=0$, then we're left with $z^2=1\implies z=\pm1$, for which $f(x,y,z)=0$. This means we have 6 critical points $(x,y,z)$ where two of the coordinates are 0, and the remaining one is either 1 or -1.

In the latter case,

$y^2z^2=x^2z^2=-\lambda\implies x^2=y^2\implies x=\pm y$

$x^2z^2=x^2y^2=-\lambda\implies z^2=y^2\implies y=\pm z$

$y^2z^2=x^2y^2=-\lambda\implies x^2=z^2\implies z=\pm x$

If $x=y=z$, then

$3x^2=1\implies x=y=z=\pm\dfrac1{\sqrt3}$

A similar thing happens in all the other cases, which leads us to getting 8 possible critical points, $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$; at each of these points, we have $f(x,y,z)=\dfrac1{27}$.

To summarize: $f(x,y,z)$ has a maximum value of 1/27 and a minimum value of 0.

Answer 2

The extreme values of a function are the minimum and the maximum values of the function.

The minimum and the maximum values are 0 and 1/27, respectively.

The function is given as:

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

$\mathbf{x^2 + y^2 +z^2 = 1}$

Subtract 1 from both sides of $\mathbf{x^2 + y^2 +z^2 = 1}$

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

Using Lagrange multiplies, we have:

$\mathbf{L(x,y,z,\lambda) = f(x,y,z) + \lambda(0)}$

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

$\mathbf{L(x,y,\lambda) = x^2y^2z^2 + \lambda(x^2 + y^2 +z^2 - 1)}$

Differentiate

$\mathbf{L_x = 2xy^2z^2 + 2\lambda x}$

$\mathbf{L_y = 2x^2yz^2 + 2\lambda y}$

$\mathbf{L_z = 2x^2y^2z + 2\lambda z}$

$\mathbf{L_{\lambda} =x^2 + y^2 +z^2 -1}$

Equate to 0

$\mathbf{2xy^2z^2 + 2\lambda x = 0}$

$\mathbf{2x^2yz^2 + 2\lambda y = 0}$

$\mathbf{2x^2y^2z + 2\lambda z = 0}$

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

Factorize the above expressions

$\mathbf{2xy^2z^2 + 2\lambda x = 0}$

$\mathbf{ 2x(y^2z^2 + \lambda) = 0}$

$\mathbf{ 2x = 0\ or\ y^2z^2 + \lambda = 0}$

$\mathbf{x = 0\ or\ y^2z^2 =- \lambda}$

$\mathbf{2x^2yz^2 + 2\lambda y = 0}$

$\mathbf{2y(x^2z^2 + \lambda) = 0}$

$\mathbf{ 2y = 0\ or\ x^2z^2 + \lambda = 0}$

$\mathbf{y = 0\ or\ x^2z^2 =- \lambda}$

$\mathbf{2x^2y^2z + 2\lambda z = 0}$

$\mathbf{2z(x^2y^2 + \lambda) = 0}$

$\mathbf{2z= 0\ or\ x^2y^2 + \lambda = 0}$

$\mathbf{z= 0\ or\ x^2y^2 =- \lambda }$

So, we have:

$\mathbf{x = 0\ or\ y^2z^2 =- \lambda}$

$\mathbf{y = 0\ or\ x^2z^2 =- \lambda}$

$\mathbf{z= 0\ or\ x^2y^2 =- \lambda }$

Because

-λ = -λ

The above expressions become:

$\mathbf{x = y = z = 0}$

$\mathbf{x^2z^2 =y^2z^2 }$

$\mathbf{x^2 = y^2}$

$\mathbf{x = \pm y}$

Also, we have:

$\mathbf{x^2y^2 =x^2z^2 }$

$\mathbf{y^2 =z^2 }$

$\mathbf{y =\pm z}$

Also:

$\mathbf{y^2z^2 =x^2y^2}$

$\mathbf{z^2 =x^2}$

$\mathbf{z =\pm x}$

So, we have:

$\mathbf{x = \pm y}$      

$\mathbf{y =\pm z}$

$\mathbf{z =\pm x}$

This means that:

$\mathbf{x = y = z}$

Recall that: $\mathbf{x^2 + y^2 +z^2 = 1}$

So, we have:

$\mathbf{x^2 + x^2 + x^2 = 1}$

$\mathbf{3x^2 = 1}$

Divide through by 3

$\mathbf{x^2 = \frac13}$

Take square roots of both sides

$\mathbf{x = \pm \frac1{\sqrt3}}$

So, we have:

$\mathbf{x =y = z= \pm \frac1{\sqrt3}}$

So, the critical points are:

$\mathbf{x = y = z = 0}$

$\mathbf{x =y = z= \pm \frac1{\sqrt3}}$

Substitute the above values in $\mathbf{f(x,y,z) = x^2y^2z^2}$

$\mathbf{f(0,0,0) = 0^2 \times 0^2 \times 0^2 = 0}$

$\mathbf{f(\pm \frac1{\sqrt3},\pm \frac1{\sqrt3},\pm \frac1{\sqrt3}) = (\pm \frac1{\sqrt3})^2 \times (\pm \frac1{\sqrt3})^2 \times (\pm \frac1{\sqrt3})^2 = \frac{1}{27}}$

Considering the above values, we have:

$\mathbf{Minimum= 0}$

$\mathbf{Maximum= \frac{1}{27}}$

Hence, the minimum and the maximum values are 0 and 1/27, respectively.

Read more about extreme values at;

brainly.com/question/1286349