Let 0 < α < β < γ ∈ R, the reals. Let K = { (x, y, z) : xβ + yβ + zβ = 1, and x, y, z ≥ 0} ∈ R3. Define f : K → R by f(x, y, z) = xα + yβ + zγ. At what points of K does f assume its maximum and minimum values?
Solution
Clearly for any (x, y, z) in K we have x, y, z ≤ 1. Hence xα ≥ xβ. In fact xα - xβ is zero at x = 0, increases to a maximum and then reduces to zero again at x = 1. Differentiating, we find the maximum is at x = (α/β)1/(β-α). Similarly zγ - zβ is non-negative, with a minimum at z = (β/γ)1/(γ-β).
To achieve a maximum for f(x, y, z), any non-zero value of z is unhelpful, whilst y is neutral, so the maximum is achieved at x = (α/β)1/(β-α), y = (1 - (α/β)β/(β-α))1β, z = 0. [The value of y is chosen so that xβ + yβ = 1.] Similarly, to achieve a minimum, we must take x = 0, y = (1 - (β/g)β/(g-β))1/β, z = (β/γ)1/(γ-β).
© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001