Minimizing a summation? I have absolutely no idea how to approach this problem. I've been looking through notes, and I think I missed this when my professor discussed this in class.
$$
\text{Consider the data}\\
i\: x_i\: y_i\\
1\:2\:1\\
2\:3\:2\\
3\:3\:3\\
4\:4\:6\\
5\:5\:5\\
\text{As discussed in class, compute the line } y=p(x)mx+b \text{ that minimizes}\\
F(m,b) = \sum_{i=0}^{5}(y_i-(mx_i+b))^2 \\
\text{You find that}\\
m=\text{____}\\
b=\text{____}\\
F(m,b)=\text{____}
$$
We've been working on maximizing and minimizing functions by both partial derivatives testing for critical points and using lagrange multipliers to find max and min values, but I have no idea what to do with this to get my initial equations to work with. Can someone give me some kind of hint or instructions on what to do?

Let our best fit line be described as $mx + b$
We would hope that we could solve for $m$ and $b$ without any contradictions using the system of equations:
$\begin{cases} 2m+b = 1\\
3m + b = 2\\
3m + b = 3\\
4m + b = 6\\
5m + b = 5\end{cases}$
Which can be expressed as the following matrix equation:
$\begin{bmatrix} 2 & 1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\begin{bmatrix}m\\b\end{bmatrix} = \begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
You will unfortunately find that this system is inconsistent and no exact solution can be found ($3m+b=2$ and $3m+b=3$ simultaneously is impossible).  However, a least squares fit can be found.  To do so, multiply on the left by the transpose of the first matrix.
$\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}2&1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\begin{bmatrix}m^*\\b^*\end{bmatrix} = \begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
This system, in the form $A^TAx=A^Tb$, will be consistent and the solution (or solution set in the case of infinitely many solutions) will be the best fit line.  Solve the system either by row-reduction, or if $A^TA$ is invertible as $x=(A^TA)^{-1}A^Tb$, or using your other favorite technique.
(Note: it is as mentioned possible for infinitely many possibilities to occur, in particular when $(A^TA)$ is non-invertible.)
In this case it will be invertible, so we take the easy route and complete the problem as:
$\begin{bmatrix}m^*\\b^*\end{bmatrix}=\left(\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}2&1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\right)^{-1}\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
$=\begin{bmatrix}\frac{5}{26}&-\frac{17}{26}\\ -\frac{17}{26}&\frac{63}{26}\end{bmatrix}\begin{bmatrix}66\\17\end{bmatrix}=\begin{bmatrix}\frac{41}{26}\\-\frac{51}{26}\end{bmatrix}$
Giving us that our best fitting line will be of the form $\frac{41}{26}x-\frac{51}{26}$

Alternatively, approaching via another method, expand $F(m,b)$ and try to calculate its minimum value using derivatives.
$F(m,b)=(1-2m-b)^2 + (2-3m-b)^2 + (3-3m-b)^2 + (6-4m-b)^2 + (5-5m-b)^2$
$=5b^2+34bm-34b+63m^2-132m+75$
We try to calculate critical points where $F_{m}=F_{b}=0$, and test $D=F_{m,m}F_{b,b} - F_{m,b}^2$ to see if they are maximums, minimums, or saddle points.
$F_m = 34b + 126m - 132$
$F_b = 10b + 34m - 34$
Setting each equal to zero, we see that each forms a line in $\mathbb{R}^2$ (and they are not parallel) and so there will be only one critical point.  Find the intersection of the line and verify that it is indeed a minimum.
These intersect at the point $m=\frac{41}{26}, b=-\frac{51}{26}$.  After some tedious arithmetic, we confirm that $D>0$ and $F_{mm}>0$, implying that this point is indeed a minimum for the multivariable function $F(m,b)$.
Giving us that our best fitting line will be of the form $\frac{41}{26}x-\frac{51}{26}$

(double checked, both methods did in fact yield the same answer)
