Question

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

Answer 1

`17/48`

Explanation:

Forming the lagrangian

`L(x,y,z,lambda)=f(x,y,z)+lambda g(x,y,z)`

with

`g(x,y,z)=4x^2+y^2-z=0`

The stationary points are computed solving

`grad L(x,y,z,lambda)=0`

or

`{(1+8lambda z=0),(2 + 2 lambda y=0),(3 + lambda=0),(4 x^2 + y^2 - z=0):}`

and solving for `(x,y,z,lambda)` we obtain

`{(x=1/24),(y=1/3),(z=17/144),(lambda=-3):}`

and the maximum value is

`17/48`

NOTE: To qualify the stationary point it is necessary to form

`(f @ g) (x,y)= x+2y -3(4x^2+y^2)`

and then calculate

`H=grad^2(f @ g) =((-24, 0),(0, -6))`

As we can observe, `H` has negative eigenvalues indicating that the found solution represents a maximum.

Answer 2

Use a Lagrange Multiplier .

Explanation:

Given: `f(x,y,z)= x+2y-3z` and the constraint function `g(x,y,z) = 4x^2+y^2-z= 0`

The Lagrange function is:

`L(x,y,z,lambda) = f(x,y,z) + lambdag(x,y,z)`

`L(x,y,z,lambda) = x+2y-3z + 4lambdax^2+lambday^2- lambdaz`

We compute the partial derivatives:

`(delL(x,y,z,lambda))/(delx) = 1+8lambdax`

`(delL(x,y,z,lambda))/(dely) = 2+ 2lambday`

`(delL(x,y,z,lambda))/(delz) =-3-lambda`

`(delL(x,y,z,lambda))/(dellambda) =4x^2+y^2-z`

Set these 4 derivatives equal to zero and then solve them as a system of equation:

`0 = 1+8lambdax" [1]"`

`0 = 2+ 2lambday" [2]"`

`0 =-3-lambda" [3]"`

`0 =4x^2+y^2-z" [4]"`

Solve equation [3] for `lambda`:

`lambda = -3`

Substitute -3 for `lambda` into equation [1]:

`0 = 1+8(-3)x" [1]"`

`x = 1/24`

Substitute -3 for `lambda` 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.