How to find the maximum value of the function f(x,y,z)= x+2y-3z subject to the constraint z=4x^2+y^2?

buscandoaireka8u

buscandoaireka8u

Answered question

2023-02-18

How to find the maximum value of the function f ( x , y , z ) = x + 2 y - 3 z subject to the constraint z = 4 x 2 + y 2 ?

Answer & Explanation

Jakob Howell

Jakob Howell

Beginner2023-02-19Added 4 answers

Forming the lagrangian
L ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z )
with
g ( x , y , z ) = 4 x 2 + y 2 - z = 0
calculating and solving for the stationary points
g ( x , y , z ) = 4 x 2 + y 2 - z = 0
or
{ 1 + 8 λ z = 0 2 + 2 λ y = 0 3 + λ = 0 4 x 2 + y 2 - z = 0
and solving for ( x , y , z , λ ) we obtain
{ x = 1 24 y = 1 3 z = 17 144 λ = - 3
and the maximum value is
17 48
NOTE: To qualify the stationary point it is necessary to form
( f g ) ( x , y ) = x + 2 y - 3 ( 4 x 2 + y 2 )
and then calculate
H = 2 ( f g ) = ( - 24 0 0 - 6 )
As we can observe, H has negative eigenvalues indicating that the found solution represents a maximum.
Guadara0c2x

Guadara0c2x

Beginner2023-02-20Added 1 answers

Given: f ( x , y , z ) = x + 2 y - 3 z and the constraint function g ( x , y , z ) = 4 x 2 + y 2 - z = 0
The Lagrange function is:
L ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z )
L ( x , y , z , λ ) = x + 2 y - 3 z + 4 λ x 2 + λ y 2 - λ z
We compute the partial derivatives:
L ( x , y , z , λ ) x = 1 + 8 λ x
L ( x , y , z , λ ) y = 2 + 2 λ y
L ( x , y , z , λ ) z = - 3 - λ
L ( x , y , z , λ ) λ = 4 x 2 + y 2 - z
These 4 derivatives should be set to zero before they are solved as a system of equations:
0 = 1 + 8 λ x [1]
0 = 2 + 2 λ y [2]
0 = - 3 - λ [3]
0 = 4 x 2 + y 2 - z [4]
Solve equation [3] for λ :
λ = - 3
Substitute -3 for λ into equation [1]:
0 = 1 + 8 ( - 3 ) x [1]
x = 1 24
Substitute -3 for λ into equation [2]:
0 = 2 + 2 ( - 3 ) y
y = 1 3
Use equation [4] to find the value of z:
z = 4 ( 1 24 ) 2 + ( 1 3 ) 2
z = 17 144
f ( 1 24 , 1 3 , 17 144 ) = 1 24 + 2 3 - 51 144
f ( 1 24 , 1 3 , 17 144 ) = 17 48
Note: One cannot use the second derivative to test whether the Lagrange multiplier has given you a maximum or a minimum; the only way to determine whether the value is a local maximum is perturbation of values. I will leave that exercise to you.

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