Absolute Max/Min of a function of two variables on a set? How do you find the absolute maximum/minimum values of the function 
$f(x,y) = x^2 + y^2 - 8y + 16$
on the given set R where $R = {(x,y): x^2 + y^2 ≤ 25}$
I know the absolute maximum is 81 and minimum is 0. How exactly does this work? I have seen something about converting the inequality in the set into an equality and then plugging it back into the equation. Every way I do this seems to be wrong and my book skips way too many steps to help. Thanks. 
 A: The extreme values of this function will occur when $f_x=0=f_y$ or along the boundary. 
$f_x=0 \implies x=0$ and $f_y=0 \implies y=4$. Note that $f(0,4)=0$. Now we search for the extreme values along the boundary and see if they are less than or greater than this value we just found. Out of all of these values (the one calculated above and the ones we are about to find on the boundary), the minimum will be the absolute minimum and the maximum will be an absolute maximum.
To Deal with the extreme values along the boundary, lets let $x=5\cos \theta$ and $y=5\sin \theta$. Then substituting this into our function yields:
\begin{align}
f(5\cos\theta,5\sin\theta)&= (5\cos\theta)^2+(5\sin\theta)^2-8\sin\theta+16\\
&=25-8\sin\theta+16\\
&=41-40\sin\theta.
\end{align}
This can be seen to have a maximum of $81$ and it can be as low as $1$. Since $f(0,4)=0<1$, we conclude that $0$ is the absolute minimum and $81$ is the absolute maximum on this set.
A: Note that
$$f(x,y) = x^2+y^2-8y+16 \leq 25 - 8y +16 = 41-8y$$
by plugging in $x^2+y^2 \leq 25$.
Also, we note that the feasible region $x^2+y^2 \leq 25$ is a closed unit disc of radius 5. Thus $-5 \leq y \leq 5$. Plugging this into our previous work.
$$f(x,y) \leq 41 - 8y \leq 41 -8(-5) =81$$
So $81$ is an upper bound on $f(x,y)$ with on the set $x^2 + y^2 \leq 25$. Furthermore, we can see that $81$ is attained at the point $(0,-5)$.
For the minimum, note that
$$f(x,y) = x^2+y^2-8y+16 = x^2 + (y-4)^2 \geq 0.$$
Since it is the sum of squares. We can see that $0$ is attained when $(x,y) = (0,4)$.
A: A geometrical interpretation of this problem is that the function $ \ z \ = \ f(x, \ y) \ = \  x^2 \ + \ (y-4)^2 \ $ is a paraboloid with circular cross-sections that "opens" in the positive $ \ z-$ direction and that we wish to find the smallest and largest values of $ \ z \ $ for this surface within or on the circular cylinder $ \ x^2 \ + \ y^2 \ = \ 25 \ $ .  Because the vertex of this paraboloid is located at $ \ (x, \ y, \ z) \ = \ (0, \ 4, \ 0 ) \ $ , which is within the "interior" of the cylinder, the absolute minimum of the function is $ \ 0 \ $ .
We can consider the projections of the level curves, which are the paraboloid's cross-sections, onto the $ \ xy-$ plane, which are circles of varying radii centered on $ \ (x, \ y) \ = \ (0, \ 4) \ $ .  It is pretty clear from considering a graph that the smallest circle which is just tangent to the constraint circle $ \ x^2 \ + \ y^2 \ = \ 25 \ $ is the circle of radius $ \ 5 - 4 \ = \ 1 \ $ , and the largest circle which is also just tangent has radius $ \ 5 + 4 \ = \ 9 \ $ .  These level curves thus correspond to function values which are the radii-squared, or $ \ f(0, \ 5) \ = \  1^2 \ $ and $ \ f(0, \ -5) \ = \  9^2 \ $ .

The Lagrange-multiplier method is a bit much for this problem, but provides another way of showing the presence of both extrema on the constraint surface:
$$ 2 \ x \ = \ \lambda \ \cdot \ 2 \ x \ \ \Rightarrow \ \ 2 \ (\lambda - 1) \ x \ = \ 0 \ \ , $$
$$ 2 \ y \ - \ 8 \ = \ \lambda \ \cdot \ 2 \ y \ \ \Rightarrow \ \ 2 \ (\lambda - 1) \ y \ + \ 8 \ = \ 0 \ \ . $$
The first equation has solutions $ \ x \ = \ 0 \ $ or $ \ \lambda \ = \ 1 \ $ , the latter of which cannot be applied consistently in the second equation. So we find on the constraint circle/cylinder, $ \ x \ = \ 0 \ \ \Rightarrow \  \ y^2 \ = \ 25 \ \ \Rightarrow \ \ y \ = \ \pm 5 \ $ ; we would be quickly led again to the absolute maximum of $ \ 81 \ $ at $ \ (0, \ -5) \ $ .
The usual "critical point" analysis in the interior of the disk/cylinder gives us $ \ 2 \ x \ = \ 0 \ \ , \ \ 2 \ y \ - \ 8 \ = \ 0 \ \ , \ \ D \ = \ 2 \ \cdot \ 2 \ - \ 0^2 \ > \ 0 \ $ , which locates the minimum of the function as $ \ 0 \ $ at $ \ (0, \ 4) \ $ .
(I wasn't entirely clear from the question statement whether an inequality technique was required, so I am posting this to show what some of the other multivariate techniques tell us.)
A: $$x^2+y^2-8y+16=x^2+(y-4)^2\ge 0$$ with equality if and only if $$x=0,y=4$$
$$x^2+y^2\le 25 \Rightarrow y^2\le 25 \Leftrightarrow -5\le y\le 5$$ so we have $$-8y\le 40$$
$$x^2+y^2-8y+16\le 25+40+16=81$$ with equality if and only if $$y=-5,x=0$$
