Let [tex]\( x_1 \)[/tex] be the number of kilograms of food type [tex]\( P \)[/tex].
Let [tex]\( x_2 \)[/tex] be the number of kilograms of food type [tex]\( Q \)[/tex].
Let [tex]\( x_3 \)[/tex] be the number of kilograms of food type [tex]\( R \)[/tex].
Let [tex]\( x_4 \)[/tex] be the number of kilograms of food type [tex]\( S \)[/tex].

What is the objective function?
[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 \][/tex]

Set up the initial simplex tableau for the problem:
[tex]\[
\begin{array}{rrrrrrrr}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \\
0 & 0 & \square & \square & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.125 & 0.125 & 0 & 1 & 0 & \square \\
1 & 0.5 & 0.25 & 0 & 0 & 0 & 1 & \square \\
\hline
\square & \square & -80 & -70 & 0 & 0 & 0 & 0
\end{array}
\]

(Type an integer or decimal for each matrix element.)

To set up the initial simplex tableau for the given problem, we need to understand the structure of the tableau based on the objective function and constraints. We deal with a linear optimization problem in which we seek to maximize or minimize an objective function subject to certain constraints.

### Objective Function and Constraints

We have an objective function:

[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 \][/tex]

We will rewrite the objective function with negative coefficients because the simplex method handles maximization by converting it to a minimization form:

[tex]\[ z = 110x_1 + 90x_2 + 80x_3 + 70x_4 + 0s_1 + 0s_2 + 0s_3 \][/tex]

### Construct the Simplex Tableau

Based on the standard form and the constraints provided in the problem, we can create the initial simplex tableau.

1. First Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 1s_1 + 0s_2 + 0s_3 \leq 600
\][/tex]

2. Second Constraint Row:
[tex]\[
\begin{array}{ccccccc}
0x_1 & + & 0.5x_2 & + & 0.125x_3 & + & 0.125x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 1s_2 + 0s_3 \leq 300
\][/tex]

3. Third Constraint Row:
[tex]\[
\begin{array}{ccccccc}
1x_1 & + & 0.5x_2 & + & 0.25x_3 & + & 0x_4 \\
\end{array}
\][/tex]
[tex]\[
+ \; 0s_1 + 0s_2 + 1s_3 \leq 400
\][/tex]

4. Objective Row: We take the negative of each coefficient from the objective function:
[tex]\[
\begin{array}{cccccccc}
-110x_1 & - & 90x_2 & - & 80x_3 & - & 70x_4 & \\
\end{array}
\][/tex]
+
[tex]\[
0s_1 + 0s_2 + 0s_3 + 0 \text{(initial profit)}
\][/tex]

### Assembling the Simplex Tableau

Putting it all together, the initial simplex tableau is:

[tex]\[
\begin{array}{cccccccc}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \text{RHS} \\
\hline
0 & 0 & 0.25 & 0 & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.125 & 0.125 & 0 & 1 & 0 & 300 \\
1 & 0.5 & 0.25 & 0 & 0 & 0 & 1 & 400 \\
\hline
-110 & -90 & -80 & -70 & 0 & 0 & 0 & 0 \\
\end{array}
\][/tex]

This tableau will be used in the simplex algorithm to iteratively find the optimal solution to the problem.