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Consider the function $$f(x, y)=x^2 y -2xy^2 +4.$$ Find the absolute maxima and minima of $f$ constrained to $2x-4y+1=0$ with $x\in [-1, 1]$.

My solution: the Weierstrass theorem guarantees that the function has absolute minima and maxima. The Lagrange multipliers method has allowed me to find only the point $(-1/4, 1/8)$. My question is: how to understand if it is a maximum or a minimum? When I have two or more points, I evaluate the function in those points and the greatest value corrispond to the maximum point and the lowe to the minimum point.

In this case, how to proceed? Thank you in advance!

EDIT: Plotted function.

enter image description here

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    $\begingroup$ Perturb the solution point and see what happens to $f$. $\endgroup$ Jan 22 at 15:42
  • $\begingroup$ Done, it should be a maximum point. But it possible that the function doesn't have a minimum? It feels strange to me. $\endgroup$
    – C. Bishop
    Jan 22 at 15:43
  • $\begingroup$ It is clearly possible to have no minimum (or maximum) within your restricted domain of $x \in [-1,1]$. Think about it. Imagine $f$ is a quadratic with a maximum. $\endgroup$ Jan 22 at 15:49
  • $\begingroup$ Have you plotted the function? $\endgroup$ Jan 22 at 15:56
  • $\begingroup$ @DavidG.Stork no because I don't know how to put the constraint together. $\endgroup$
    – C. Bishop
    Jan 22 at 16:20
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Here's the surface and the constraint line:

enter image description here

Although the curvature on the constraint line is low, there is a single maximum, as is to be expected in a quadratic. The minima occur at the boundaries of the region... and Lagrangian methods do not find those.

If you're given a bounded region, you may have to check boundaries to find maxima and minima. After all, consider the trivial problem of finding the maxima and minima of $f(x)=x$ in the range $x \in [0,1]$. Even basic calculus will not yield the extrema. You must check boundaries. The same sometimes occurs with Lagrangian methods.

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  • $\begingroup$ Thank you for the answer. So, minima also exist. If Lagrangian methods doesn’t work, how to find these minima? $\endgroup$
    – C. Bishop
    Jan 23 at 6:50
  • $\begingroup$ Check boundaries. $\endgroup$ Jan 23 at 17:10
  • $\begingroup$ If evaluate in the boundaries (I mean for $x=-1$ and $x=1$), I found the points $P=(-1, -1/4)$ and $Q=(1, 3/4)$. Evaluated in $f(x, y)$, I have $f(P)= 31/8$ and $f(Q)=29/8$, so I conclude that $Q$ is a minimum. Could you please tell me if mi reasonings are true? Thank you in advance! $\endgroup$
    – C. Bishop
    Jan 24 at 7:56

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