Find the minimum and maximum values of the function subject

Find the minimum and maximum values of the function subject to the given constraint(s). f(x, y, z) = 3x + 2y + 4z, x^2 + 2y^2 4- 6z^2 = 1. f(x, y, z) = x + y + z, x^2 + y^2 + z^2 = 9, 1/4x^2 + 1/4y^2 + 4z^2 = 9.

Solution

1 ) Given that

f( x , y) = 3x + 2y + 4z subject to x² + 2y² + 6z² = 1

Let g(x , y) =   x² + 2y² + 6z² = 1

Using legrange multipliers finding maxima and minima

f(x,y) = g(x,y)

< 3 , 2 , 4 > = < 2x , 4y , 12z >

  .2x = 3 ,  .4y = 2 ,  .12z = 4

x = 3/2 , y = 2/4 , z = 4/12

x = 3/2 , y = 1/2 , z = 1/3

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

x² + 2y² + 6z² = 1

( 3/2 )² + 2( 1/2 )² + 6(1/3 )² = 1

123 / 36² = 1

123 = 36²

² = 123/36

² = 41/12

   = ± 123/6

x = 3/2 = (3/2)(1/) = (3/2)(± 6/123) = ± 9/123 = 9/123 , - 9/123

y = 1/2 = (1/2)(1/) = (1/2)(± 6/123) =  ± 3/123 = 3/123 , -3/123

z = 1/3 = (1/3)(1/) = (1/3)(± 6/123) = ± 2/123 = 2/123 ,-2/123

Substitute x , y , z in f(x,y) = 3x + 2y + 4z

= 3(9/123) + 2(3/123) + 4(2/123)

= 27/123 + 6/123 + 8/123

= 41/123

= 123 /3

Therefore,

Maximum value occurs at f( 9/123 , 3/123 , 2/123 ) = 123 /3

Minimum value occurs at f( -9/123 , -3/123 , -2/123 ) = -123 /3