0

If I maximize $x^3/3-3y^2/2+2x \;$ subject to $x-y=0$ by using the Lagrangean method and confirm the bordered Hessian condition, I get that no solution exists. This can also be seen by looking at the graph.

However, if I substitute the constraint into the objective function and treat it as an unconstrained optimization problem, then I get (1,1) as the maximum and (2,2) as the minimum. Clearly, this is incorrect, as it doesn't agree with the graph.

In this context, I want to ask when does substitution work and when does it not? Should we never simplify a problem using substitution?

PGupta
  • 609

1 Answers1

1

You're doing something wrong with the Lagrangian. Indeed, $(x,y)=(1,1)$ is a maximum and $(x,y)=(2,2)$ is a minimum as you can easily verify by plotting $f(x)=x^3/3-3x^2/2+2x$. Both methods yield this result. Since you haven't posted your derivations, it's hard for me to point out the exact mistake.

Greased Dog
  • 43
  • 4
  • I don't think my Lagrangian is wrong. Pls follow this link to see the graph. I have plotted the level curve of the objective function, the level can be changed using the slider on $k$. https://www.desmos.com/calculator/mchihtb4ji – PGupta May 12 '21 at 12:15
  • It seems you are confused about maxima/minima and global maxima/minima. The function does not have a global maximum or minimum, but the previously mentioned points are indeed (local) minima and maxima, respectively. Note that both methods don't tell us anything (without further ado) about if the maxima and minima are global or local. – Greased Dog May 12 '21 at 12:46
  • What additional steps are needed to show that global optima don't exist in this case? – PGupta May 13 '21 at 03:22
  • 1
    Under the constraint, the function is simply polynomial of degree 3. I think it's quite obvious that polynomials of odd degree never have global minima or maxima (as they will always tend to plus/minus infinity when x tends to plus/minus infinity, or the other way around) – Greased Dog May 13 '21 at 17:11