Linear Programming and graphing (Mathematics in thw Modern World) I need help with the #1 question. linear Programming is applied. 
1. Consider the recipes below:
Pancakes                           -Waffles 
3 cups Bisquick                     3 cups Bisquick 
1 cup Milk.                         2 cups Milk
2 eggs.                             2 eggs 
Serves 6.                           Serves 5

You have 24 cups of Bisquick, 18 cups of milk, and 20 eggs. If you want to feed as many people as possible, how many batches of each should you make?
I barely understand linear Programming so this confused me even more. The ingredients confuse me. I kind of know the basics of linear Programming but this one, I don't understand.  I need help please, thank you.  
 A: Firstly we define the variables: $x_1$: Multiple of pancake recipe, $x_2$: Multiple of waffle recipe.
We want to maximize the total amount of serves. Therefore the objective function is
$$\text{max} \ \ 6x_1+5x_2$$
Next we have several constraints for the ingredients. For instance, that you have 24 cups of Bisquick. The corresponding constraint is
$$3x_1+3x_2\leq 24$$
If we use the double amount  of the pancake recipe   and three times of the waffles recipe we need $3\cdot 2+3\cdot 3=15<24$ cups of Bisquick. The constraint is fulfilled. You can call this constraint the Bisquick constraint. You make similar constraints for  milk and eggs.
Finally we need the non-negativity constraint: $x_1,x_2\geq 0$. This real world problem requires that the variables are integers, but at the first step we can use the non-negativity constraint.

Update
The coefficients of the objective functions are negative in the table since it is a max problem. The first row is the objective function. For each $\leq$-problem we need a slack variable ($s_i)$. 
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \color{blue}{x_1}&x_2&s_1&s_2&s_3&RHS \\ \hline -6&-5&0&0&0&0 \\ \hline \color{red}3&3&1&0&0&24 \\ \hline 1&2&0&1&0&18\\ \hline 2&2&0&0&1&20\\ \hline \end{array}$$
This is the initial table. The pivot column is $x_1$ since $|-6|$ is larger than $|-5|$. Now we look for the minimum of the fractions of the RHS and the corresponding coefficients of colunm $x_1$.
$\min\left(\frac{24}{3},\frac{18}{1},\frac{20}{2}\right)=\min\left(8,18,10\right)=8$
That means the pivot row is the second row in the table. And therefore the first pivot element is $\color{red}3$.
