In Pablos problem via Simplex method :
The Conundrum City school board is heavily influenced by the local fruit grower's association. They have stipulated that children eat at least $7$ oranges and $5$ apples per week. Parents and teachers have agreed that eating at least $15$ pieces of fruit per week is a good thing, but school janitors argue that too much fruit makes a terrible mess, so that children should eat no more than $25$ pieces of fruit per week.
$$s = 5x + 10y$$
This linear function of $f(x, y)$ represents the grams of sugar in $x$ apples and $y$ oranges.
The problem is asking us to minimize s subject to the four linear inequalities listed above. $$x\ge 5$$ $$y\ge 7$$ $$x + y \le 25$$
$$\implies15 \leq x + y \leq 25 $$
But in Simplex method, all variables are to obey the non-negativity $x_i \geq 0$ constraint.
So let’s select $x_1=x-5$ and $x_2=y-7$; Hence,
$\implies 15 \leq (x1+x2)+12 \leq 25$
$\implies 3 \leq (x1+x2) \leq 13$
Since these equations have inequalities $\leq, \geq $ and hence NOT of the form $Mx = v$ where $M$ is some Matrix of coefficients.
We introduce slack variables $x_3 \geq 0$ and $x_4 \geq 4$.
QUESTION: I don’t understand why $x_4 \geq 4$ How 4?
$ \implies c_1:= x_1 + x_2 -x_3 = 3\ \ \ , \ \ \ \ c2:= x_1 + x_2 + x_4 = 13$
The matrix form looks like: $$ \begin{pmatrix} 1 & 1 & -1 & 0 \\ 1 & 1 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{pmatrix} = \begin{pmatrix} 3\\ 13\\ \end{pmatrix} $$
We also want to minimize sugar: $s = 5x + 10y$, but the STANDARD form of the simplex method maximizes, hence:
objective function f = $-s + (5*5 + 7*10) = -5x_1 - 10x_2 + 95$
Note: Notice that it makes no difference whether we maximize $-s$ or $-s + 95$, we choose the latter since it is a linear function of $(x_1, x_2)$
$5x_1 + 10x_2 + f =0$:
$$ \left[\begin{array}{ccccc} x_1 & x_2 & x_3 & x_4 & f \end{array}\right] $$
$$\implies \left[\begin{array}{rrrrr|r} 1 & 1 & 1 & 0 & 1 & 3 \\ 1 & 1 & 0 & 1 & 0 & 13 \\ \hline 5 & 10 & 0 & 0 & 1 & 0 \end{array}\right] $$
here, it seems simplex algo already terminated as last row has (+) coefficients $\implies x_1=0=x_2$ would be optimal. However,this does not solve constraints for positive values of slack variables {$x_3, x_4$}. So we need to add artificial variables {$x_5, x_6$}. Such that:
$$c_1 \rightarrow c_1 - x5\ \ , \ \ \ \ c_2 \rightarrow c_2 - x6$$
The above is done so that for a large positive $\alpha$ the objective function $f$ is only maximal when the artificial variables disappear = 0. Where $f - \alpha x_5 - \alpha x_6$. Taking $\alpha = 10$ we get: $$ \left[\begin{array}{ccccccc|c} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & f \end{array}\right] $$
$$ \left[\begin{array}{rrrrrrr|r} 1 & 1 & -1 & 0 & 1 & 0 & 0 & 3 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 13 \\ \hline 5 & 10 & 0 & 0 & 10 & 10 & 1 & 0 \end{array}\right] $$
After a series of EROs:
$$R'_3 = R_3 + 15R_1$$
$$R'_2 = R_3 + 15R_1$$
$$R'_3 = R_3 + 10R_2$$
We arrive at: $$ \left[\begin{array}{rrrrrrr|r} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 1 & -1 & 1 & 0 & 10 \\ \hline 0 & 5 & 5 & 0 & 5 & 10 & 1 & -15 \end{array}\right] $$
The solution in the book says {$x_2, x_3, x_5, x_6$} have zero coefficients? But looking at the matrix above I do not see zero as a coefficient for $x_2$; rest are slack and artificial, so can it assumed to be zero? Not sure about any of this, please help understand this step.
Here is a screenshot of this section in the book, and my edits of my interpretation of these steps:
