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I am confused as to whether to use the standard hessian or bordered hessian for the following problem.

$f(x,y,z) = (x+1)^2+(y+1)^2+z^2 \text{ subject to } x+y=(x-y)^2 \text{ and } z-x-y=1$

We are told that $(0,0,1)$ satisfies the first order conditions. Show that $(0,0,1)$ is the unique global min.

We were previously given a similar problem (which was not constrained) that said there is exactly one stationary point and to show that it is the unique global max. The standard hessian was all we needed for this problem.

I was initially inspired to use the bordered hessian for the above problem since it is constrained, but now I am unsure because of the latter example.

atd31
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1 Answers1

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When you have an optimization problem with constraints, you must use the bordered hessian. The standard hessian simply will not give you the correct answer.

Example: Let's look at a simple example. Find the extrema of $f(x,y)=x^2+y^2$ restricted to the ellipse $g(x,y)=4x^2+y^2-1=0$. It easy to see that there are $2$ maxima and $2$ minima. However, the ordinary hessian (and second derivatives) in the four extrema will be positive. So from the standard hessian, you cannot deduce the correct answer.

For completeness, I want to mention that if you only want to know global extrema, it is not always necessary to use the bordered hessian. If the contraints form a closed and bounded curve, surface, or higher dimensional surface, then it suffices to calculate the function values at the extrema and pick the points where the values are maximal/minimal.

In this question however, the curve determined by the constraints is not bounded (it's a parabola), so you would still need to check whether the function values don't get smaller when "going to infinity along the parabola".

Ernie060
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