Lagrange min max Find the max/min of $ \ f(x,y) \ = \ x^2+y^2-12x+16y \ $ .
Is my proof completely right?
Solution:
By Lagrange for $x^2+y^2=25$ we have 
$$ f_x \ = \ 2x-12 \ = \ \lambda \ \cdot \ 2x \ \ , \ \ f_y \ = \ 2y+16 \ = \ \lambda \ \cdot \ 2y \ \ . $$
From here, $ \ x \ = \ \frac{6}{1-\lambda} \  $ and $ \ y \ = \ \frac{-8}{1-\lambda} \ $ .  Plugging in the equation $x^2+y^2=25$ give us two answers for $ \ \lambda \ $  which give us the pairs $ \ (-3, \ 4) \ $ and $ \ (3, \ -4) \ $ and $ \ \lambda \ \neq \ 1 $ .
However the problem is not solved!  The constraint is   $ \ x^2 \ + \ y^2 \ \le \ 25 \ $ .  So I will find the local min/max and compare them to the results of Lagrange.  First it is easy to see that each $ \ x \ $ and $  \ y \ $ are in $ \ [-5,5] \ $ . The first derivative with respect to $ \ x \ $ is equal to $ \ f_ x \ = \ 2x-12 \ $ , but $ \ x \ $ is in $ \ [-5,5] \ $ .  We know that if $ \ f_ x \ > \ 0  \ $ the function grows and if $ \ f_ x \ < \ 0  \ $ it becomes smaller.  So this means that with respect of $ \ x \ $ , the function has max when $ \ x = 5 \ $ and min when $ \ x = -5 \ $ or we have pairs $ \ (-5,0) \ $ and $ \ (5;0) \ $ .  Doing the same thing for the derivative   $ \ f_y \ $ , we get also two answers for min/max and they are $ \ -5 \ $ and $ \ 5 \ $ ; then we have pairs $ \ (0,-5) \ $ and  $ \ (0,5)\ $.  
Now our answer is the minimum/maximum value when we plug in the following pairs from all cases $ \ (-5,0)  \ , \  (5,0) \ , \   (0, -5) \ , \ (0,5)  \ ,  \ (-3,4) \ , \ (3, -4) \ $ .  By testing, the final result is the pairs $ \ (-3,4) \ $ and $ \ (3,-4) \ $ .
 A: Your extremal points are correct, but you   made rather more work for yourself than you needed to.  The Lagrange-multiplier method does tell us about the extremal points on the constraint circle.  So we find a minimum value $ \ -75 \ $ at $ \ (-3, \ 4) \ $ and a maximum of  $ \ 125 \ $ at $ \ (3, \ -4) \ $ .
You are also correct in saying that we are not finished, as we must also investigate the "interior" of the disk $ \ x^2 \ + \ y^2 \ < \ 25 \ $ .  For that, we would use "critical point" analysis, which really proceeds in the same way that we use for functions of one variable:  we have $ \ f_x \ = \ 2x \ - \ 12 \ = \ 0 \ \ ,$ $ \ \ f_y \ = \ 2y \ + \ 16 \ = \ 0 \ $ , which tells us that there is a critical point for $ \ f(x, \ y) \ $ at $ \ (6, \ -8) \ $ .  But this point is completely outside the constraint circle, so there is no critical point within the disk.  So our examination is complete -- we have found the absolute maximum and minimum of our function within the constraint disk and on its boundary.
(Using an argument based on your discussion of the first partial derivatives, you would perhaps want to say something more like this.  We have
$$ -5 \ \le \ x \ \le \ 5 \ \ , \ \ -5 \ \le \ y \ \le \ 5 \ $$
$$ \Rightarrow \ \ 2 \ (-5) \ - \ 12 \ \le \ f_x \ \le \ 2 \ (5) \ - \ 12 \ \ , \ \ 2 \ (-5) \ + \ 16 \ \le \ f_y \ \le \ 2 \ (5) \ + \ 16 $$
$$ \Rightarrow \ \   -22 \ \le \ f_x \ \le \   -2 \ \ , \ \  6 \ \le \ f_y \ \le \ 26 \ \ . $$
Hence, there are no points within the square $ \ [-5, \ 5] \ \times \ [-5, \ 5] \ $ , which contains the constraint circle, for which the first derivatives are zero, and thus there are no critical points within the interior of the circle.)
$ \ \ $
There is a geometrical interpretation for this problem.  Considering the function $ \ z \ = \ f(x,y) \ = \ x^2 \ + \ y^2 \ - \ 12x \ + \ 16y \ = \ (x-6)^2 \ + (y + 8)^2 \ - \ 100 \ $ as describing a surface in three dimensions, this is a paraboloid with circular cross-sections "opening" in the positive $ \ z-$ direction with its vertex at $ \ (x, \ y, \ z) \ = \ (6, \ -8, \ -100 \ ) $ .  We are looking for the extremal values of $ \ z \ $ where this paraboloid passes on or within the circular cylinder $ \ x^2 \ + \ y^2 \ = \ 25 \ $ .
From this description, we see that the minimal value of the function, $ \ -100 \ $ does not fall within the constraint region.  Since the "height" of the surface relative to the $ \ xy-$ plane only increases as we "move away" from the vertex, we   expect that there can be no extrema within the constraint cylinder.  So the extremal points will only be found on the surface of the cylinder, $ \ x^2 \ + \ y^2 \ = \ 25 \ $ .
The level curves of our function are circles centered on $ \ (x, \ y) \ = \ (6, \ -8) \ $ .  The Lagrange-multiplier method locates points where these level curves are just tangent to the constraint circle, which you found to lie at $ \ (-3, \ 4) \ $ and $ \ (3, \ -4) \ $ .  The function $ \ f(x, \ y) \ $ can be understood as a "distance-squared" function of points measured from $ \ (6, \ -8) \ $ ; the closest  point on the constraint circle is $ \ (3, \ -4) \ $ , the farthest, $ \ (-3, \ 4) \ $ .
If we take the ratio between the Lagrange equations, with $ \ \lambda \ \neq \ 1 \ $ , we obtain
$$ \ \frac{(\lambda \ - \ 1) \ \cdot \ 2 \ y}{(\lambda \ - \ 1) \ \cdot \ 2 \ x} \ = \ \frac{8}{-6} \ \ \Rightarrow \ \ \frac{y}{x} \ = \ -\frac{4}{3} \ \ . $$
[We can neglect the case of $ \ \lambda \ = \ 1 \ $ , as it cannot be applied consistently in the two equations.]
This tells us that the two extremal points lie on the line $ \ y \ = \ -\frac{4}{3} \ x \ $ , marked in light blue on the graph below.  Since the constraint circle is symmetric about the origin, and the function to be extremized is radially symmetric about its center, we would expect that the extremal points will lie on a line through the origin, diametrically opposite one another on the constraint circle.

Here is a similar problem, but with the minimum of the function falling within the constraint circle.
A: Let $ g(x,y)= x^2+y^2-25$.
The Lagrange multiplier and required points can in other words be evaluated by:
$$ \lambda = \dfrac{f_x}{g_x}=\dfrac{f_y}{g_y}  $$
Second order derivatives ( 6 in total) can be used to further establish min/max  as with the single independent variable.
