proving that a point is a global minimum

I have this cost function

$$ f(x,y,z)=0.5 \left(x^2+y^2+0.1(z^2)\right)+0.55z $$

such that

$$x,y,z>0$$

$$x+y+z=1$$

I was given the point $(0.5,0.5,0)$ as a local minimum. How to I go along to prove/disprove this is also a global minimum.

Lagrange multipliers just gave me a max point.

Thank you.

1 Answer

The differential of your function:

$d(0.55 z + 0.5 (x^2 + y^2 + 0.1 z^2)) = x dx + y dy + (0.55 + 0.1 z) dz$

A global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the domain (This is the Extreme Value Theorem). Local maxima and minima occur wherever the derivative of the function at that location is zero. These are "critical points" because they either have a derivative of zero or are a boundary point and are therefore nondifferentiable.

Therefore, the next step is to find the critical points.

The critical points of this function are:

Where $x=0$, $y=0$, and $(0.55 + 0.1 z)=0 \implies z=-5.5$ (not valid because z must be > 0)

and

The boundary points: $x=0$, $\space \space$ $y=0$, $\space \space$ $z=0$, $\space \space$$x=1$, $\space \space$$y=1$, $\space \space$ $z=1$ (every combination thereof is a boundary point; there are $2^3=8$ points total.)

We must check these points by evaluating $f(x,y,z)$ at each point to determine the minimum.

Checking all points, we find that the minimum is at $(x,y,z)=(0,0,0)$.

Additional reading:

According to the Extreme Value Theorem, if a function is continuous on a closed interval, a maximum and minimum exist. (wikipedia)

see also How to prove local minima are global?

View the answer to this question on wolframalpha.com (as kind of a hack I put in the z term as abs(z))