<s>[INST] Use Lagrange Multipliers to find the absolute extrema Use Lagrange Multipliers to find the absolute extrema (if any) of:
$f(x,y) = 4x^2 + 9y^2$; subject to $2x +3y = 6$.
Using Lagrange I end up with one point: $(\frac{3}{2}, 1)$
I'm just not sure how to show if that point is a max or a min?
 [/INST] Another way to check the nature of the extremum in this situation is to substitute the constraint into the function.  Thus, with $ \ 3y \ = \ 6 \ - \ 2x \ $ , we have
$$ f(x,y) \ = \ 4x^2 \ + \ 9y^2 $$
$$ \rightarrow \ \ \phi(x) \ = \ 4x^2 \ + \ (6 \ - \ 2x)^2 \ = \ \ 4x^2 \ + \ 36 \ - \ 24x \ + \ 4x^2 \ = \ 4 \ ( \ 2x^2 \ - \ 6x \ + \ 9 \ ) $$
$$ = \ 4 \ ( \ 2 \ [ \ x^2 \ - \ 3x \ ] \ + \ 9 \ ) \ = \ 4 \ \left( \ 2 \ [ \ x^2 \ - \ 3x \ + \ \frac{9}{4} \ ] \ + \ 9 \ - \ \frac{9}{2} \ \right) $$
$$ = \ 4 \ \left( \ 2 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ \frac{9}{2} \ \right) \ = \ 8 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ 18 \ \ . $$
We have thereby shown, since this function describes an "upward-opening" parabola, that the minimum of $ \ \phi(x) \ $ or $ \ f(x,y) \ $ is 18, which occurs at $ \ x \ = \ \frac{3}{2} \ \Rightarrow \ y \ = \ 1 \ . $ 

A geometrical interpretation of the problem is that, by using the Lagrange-multiplier method, we are looking for level curves of the function $ \ f(x,y) \ $ which are just tangent to the constraint "curve", which is the line $ \ 2x \ + \ 3y \ = 6 \ $ .  The level curves  $  \ 4x^2 \ + \ 9y^2 \ = \ C \ $ are concentric ellipses, only one of which is large enough to "touch" the constraint line without passing through it.  That ellipse is $  \ 4x^2 \ + \ 9y^2 \ = \ 18 \ $ ; ellipses with larger values of $ \ C \ $ cross the line at two points and can be made arbitrarily large.  Thus, the Lagrange-multiplier method locates the absolute minimum of $ \ f(x,y) \ $ ; there is no absolute maximum.
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