Find the minimum value of the function f(x,y,z) = (3x + y - 2z)/sqrt(x^2 + y^2 + z^2)?

Converting to spherical coordinates:
f(θ, φ) = [3(ρ sin φ cos θ) + (ρ sin φ sin θ) - 2(ρ cos φ)]/ρ
...........= (3 cos θ + sin θ) sin φ - 2 cos φ.

(Note that f is independent of ρ!)

Here, observe that θ is in [0, 2π] and φ is in [0, π].
(Consequently, f attains its extreme values.)
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First, we find critical points inside the region for θ and φ.

Computing the first partial derivatives,
f_θ = (-3 sin θ + cos θ) sin φ
f_φ = (3 cos θ + sin θ) cos φ + 2 sin φ.

Set these equal to 0:
(-3 sin θ + cos θ) sin φ = 0
(3 cos θ + sin θ) cos φ + 2 sin φ = 0.

The first equation implies that
-3 sin θ + cos θ = 0, or sin φ = 0
==> θ = tan⁻¹(1/3) or tan⁻¹(1/3) + π, or φ = 0 or π.

If θ = tan⁻¹(1/3), then the second equation yields
(3 * 3/√10 + 1/√10) cos φ + 2 sin φ = 0, via 'sohcahtoa'
==> √10 * cos φ + 2 sin φ = 0
==> tan φ = -√10/2
==> φ = tan⁻¹(-√10/2) + π.

If θ = tan⁻¹(1/3) + π, then the second equation yields
(3 * -3/√10 - 1/√10) cos φ + 2 sin φ = 0, via 'sohcahtoa'
==> -√10 * cos φ + 2 sin φ = 0
==> tan φ = √10/2
==> φ = tan⁻¹(√10/2).

If φ = 0 or π, then the second equation yields
(3 cos θ + sin θ) * ±1 + 0 = 0
==> tan θ = -1/3.
==> θ = π + tan⁻¹(-1/3) or 2π + tan⁻¹(-1/3).

Testing the critical points:
f(θ, φ) = (3 cos θ + sin θ) sin φ - 2 cos φ

f(tan⁻¹(1/3), tan⁻¹(-√10/2) + π) = √14
f(tan⁻¹(1/3) + π, tan⁻¹(√10/2)) = -√14
f(π + tan⁻¹(-1/3), 0) = -2
f(2π + tan⁻¹(-1/3), 0) = -2
f(π + tan⁻¹(-1/3), π) = 2
f(2π + tan⁻¹(-1/3), π) = 2.
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Checking the boundaries of the region:
f = (3 cos θ + sin θ) sin φ - 2 cos φ

(i) θ = 0 or 2π:
==> f = 3 sin φ - 2 cos φ
.........= √(3^2 + (-2)^2) sin(φ - t) for some t (which can be found if needed).
.........= √13 sin(φ - t).

This has extreme values -√13 and √13.

(ii) φ = 0
==> f = -2 (constant).

(iii) φ = π
==> f = 2 (constant).
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Overall, the extreme values of f are -√14 and √14.
(Double checked on Wolfram Alpha.)
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I hope this helps!