Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z) = 8x + 8y + 4z; 4x2 + 4y2 + 4z2 = 36 maximum value minimum value

Answer 1

Answer:

The maximum value= 36

Minimum value = - 36

Step-by-step explanation:

Given that

f(x, y, z) = 8 x + 8 y + 4 z

h(x,y,z)=4 x² + 4 y² + 4 z² - 36

From Lagrange multipliers

Δf = λ Δh

Δf = < 8 ,8 , 4>

Δh = < 8 x ,8 y  , 8 z>

Δf = λ Δh

So

< 8 ,8 , 4> = < 8  λ x ,8 λ y  , 8 λ z>

8 = 8  λ x                     -------------1

8 = 8 λ y                      ----  ------2

4 = 8 λ z                    ----------------3

From equation 1 ,2 and 3

Now by putting the value of x,y and z in the following equation

4 x² + 4 y² + 4 z² = 36

$4\times \dfrac{1}{\lambda^2 }+4\times \dfrac{1}{\lambda^2 }+4\times \dfrac{1}{(2\lambda)^2 }=36$

$\dfrac{4}{\lambda^2 }+ \dfrac{4}{\lambda^2 }+ \dfrac{1}{\lambda^2 }=36$

So the value of λ is

$\lambda =\pm \dfrac{1}{2}$

When λ = 1/2

x = 1 / λ   , y=1 / λ   ,  z= 1 /2 λ

x= 2 , y = 2 , z=1

So

f(x, y, z) = 8 x + 8 y + 4 z

f(2, 2, 1) = 8 x 2 + 8 x 2 + 4 x 1

f(2, 2, 1) =36

When λ = - 1/2

x = 1 / λ   , y=1 / λ   ,  z= 1 /2 λ

x= - 2 , y = - 2 , z= - 1

So

f(x, y, z) = 8 x + 8 y + 4 z

f(-2, -2, -1) = 8 x (-2) + 8 x (-2) + 4 x (-1)

f(-2, -2, -1) = - 36

The maximum value= 36

Minimum value = - 36

Answer 2

Answer:

Maximum value of f(x,y,z)=36 at (2,2,1)

Minimum value of f(x,y,z)=-36 at (-2,-2,-1)

Step-by-step explanation:

We are given that

$f(x,y,z)=8x+8y+4z$

$g(x,y,z)=4x^2+4y^2+4z^2=36$

We have to find the extreme values of the function using Lagrange multipliers.

$f_x(x,y,z)=8$

$f_y(x,y,z)=8$

$f_z(x,y,z)=4$

$g_x(x,y,z)=8x$

$g_y(x,y,z)=8y$

$g_z(x,y,z)=8z$

$f_x=\lambda g_x$

$8=8x\lambda$

$x=\frac{1}{\lambda}$

$f_y=\lambda g_y$

$8=8y\lambda$

$y=\frac{1}{\lambda}$

$f_z=\lambda g_y$

$4=8z\lambda$

$z=\frac{1}{2\lambda}$

Substitute the values in g(x,y,z)

$4(\frac{1}{\lambda})^2+4(\frac{1}{\lambda})^2+4(\frac{1}{2\lambda})^2=36$

$\frac{4}{\lambda^2}+\frac{4}{\lambda^2}+\frac{1}{\lambda^2}=36$

$\frac{9}{\lambda^2}=36$

$\lambda^2=\frac{9}{36}=\frac{1}{4}$

$\lambda=\pm\frac{1}{2}$

Substitute $\lambda=\frac{1}{2}$

$x=2,y=2,z=1$

Substitute $\lambda=-\frac{1}{2}$

$x=-2,y=-2,z=-1$

Now, $f(2,2,1)=8(2)+8(2)+4(1)=16+16+4=36$

$f(-2,-2,-1)=8(-2)+8(-2)+4(-1)=-16-16-4=-36$

Maximum value of f(x,y,z)=36 at (2,2,1)

Minimum value of f(x,y,z)=-36 at (-2,-2,-1)