What I tried:
3a1 Make $c_3$ greater than the current $z_3$ so any value greater than 20/3?
3a2 I just do EROs to make the $x_3$ column into $[0, 0, 1]^T$ ?
3b $c_1 \ge 10$? idk
3c If we have 630 instead of 600, then I think (instead of re-solving), I think we make use of:
$z' = z + (z_j - c_j)(x_j)$
?
How can I approach this?
3d I think of rewriting the LP problem with $x_3=10$:
$$z=10x_1+6x_2$$
s.t.
$$x_1 + x_2 (+ x_4) \le (=) 90$$
$$5x_1 + 2x_2 (+ x_5) \le (=) 275$$
$$x_1 + x_2 (+ x_6) \le (=) 120$$
If asked only for change in obj function value then, we can just do this I believe:
$z' = z + (z_j - c_j)(x_j) = 2200/3 + (8/3)(10)$
Is that right?
3e $\begin{bmatrix} 6\\ 10\\ 0 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 4\\ 3 \end{bmatrix}$?
From Chapter 2 here.
a) You are correct, you need $$c_3>20/3$$
b) In the optimal tableau, replace $c_1=10$ by $10+\Delta$, and compute the new reduced costs $z_j-c_j$. You want them to remain positive in order for the tableau to remain optimal, and this gives you $$ -4<\Delta<5\quad \Rightarrow\quad 6<c_1<15 $$
c) Replace $600$ by $600+30$ in the second constraint. The extra $30$ will be processed like $x_5$, so you will get an additional $30z_{x_5}$ in your objective function, in other words the objective function is increased by 20 units: $$ z'=z+20 $$
d) You are almost correct, but this will decrease the objective function: $$ z'=z-\frac{8}{3}10 $$