Depot D1 holds 35 crates and depot D2 holds 45 crates…

Step 1. The model. D2 sends $20 - x$ crates to S1 and $40 - y$ to S2. The cost is $Z = 8x + 6y + 6(20 - x) + 9(40 - y) = 2x - 3y + 480$ rupees. The rules are $x + y \le 35$ for D1, $(20 - x) + (40 - y) \le 45$ which is $x + y \ge 15$ for D2, $x \le 20$, $y \le 40$, $x \ge 0$ and $y \ge 0$.

Step 2. The costs. At $(15, 0)$, $Z = 510$. At $(20, 0)$, $Z = 520$. At $(20, 15)$, $Z = 40 - 45 + 480 = 475$. At $(0, 35)$, $Z = -105 + 480 = 375$. At $(0, 15)$, $Z = -45 + 480 = 435$. The smallest is $375$.

Step 3. Say it. D1 sends $0$ crates to S1 and $35$ to S2. D2 sends $20$ to S1 and $40 - 35 = 5$ to S2. The total cost is $375$ rupees.