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
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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)`
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