# Use Lagrange multipliers to find the indicated extrema: Maximize f(x,y,z)=x+y+z subject to x^2 +y^2 +z^2 =1

mathsworkmusic | (Level 2) Educator

Posted on

We want to maximise `f(x,y,z) = x+y+z`

subject to the constraint `C(x_0,y_0,z_0)=1` where `C(x,y,z) = x^2+y^2+z^2`

The method of lagrange multipliers involves simultaneously solving

1) `(delf)/(delx) -lambda(delC)/(delx) =0`

2) `(delf)/(dely) -lambda(delC)/(dely) =0`

3) `(delf)/(delz) -lambda(delC)/(delz) = 0`

where `x^2+y^2+z^2 = 1`

`implies` simultaneously solving

1)  `1 - 2lambdax = 0`   2) `1-2lambday =0`  and 3) `1-2lambdaz = 0`

`implies`  `x = y = z= 1/(2lambda)`

The constraint, then, is that

`(1/(4lambda^2)) + (1/(4lambda^2)) + (1/(4lambda^2)) = 1`

ie that `3/(4lambda^2) = 1`  `implies` `lambda = +-sqrt(3/4)`

Now, if `lambda` is negative then `f(x,y,z)` is at its minimum. If `lambda` is positive on the other hand, `f(x,y,z)` is at its maximum.

Therefore `f(x,y,z) = x+y+z`  is maximised when `lambda = sqrt(3/4)`

Then, `x=y=z= 1/(2sqrt(3/4)) = 1/sqrt(3)`

and `f(x,y,z) = 3/sqrt(3) = sqrt(3)`

NB we could have done this using symmetry, ie `f` is maximised when `x=y=z`. This is when `3x^2 = 1` `implies` `x = sqrt(1/3) = 1/sqrt(3)`

The maximum value of f(x,y,z) is `sqrt(3)`