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

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

So the value of λ is

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

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

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

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

Substitute

Substitute

Now,

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

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