To solve this optimization problem, we first need to define the decision variables, constraints, and the objective function. Let's denote:
- \(x\) as the number of laptops produced,
- \(y\) as the number of tablets produced.

The objective is to maximize profit. Given that each laptop gives a profit of $200 and each tablet gives a profit of $160, the objective function can be written as:
\[ \text{Maximize} \quad 200x + 160y \]

There are two main constraints:
1. **Manufacturing Time Constraint**: Each laptop takes 20 minutes, and each tablet takes 15 minutes. The total manufacturing time available is 1200 minutes.
\[ 20x + 15y \leq 1200 \]
2. **Silicon Availability Constraint**: Each laptop requires 3 units of silicon, and each tablet requires 2 units. There are 150 units of silicon available.
\[ 3x + 2y \leq 150 \]
Additionally, there's a constraint that the company must make at least 30 laptops:
\[ x \geq 30 \]

Since we cannot produce negative quantities of laptops or tablets, we also have non-negativity constraints:
\[ x \geq 0 \]
\[ y \geq 0 \]

Given these definitions, we can now write the Gurobi code to solve this linear programming problem.

```python
from gurobipy import *

# Create a new model
m = Model("Laptop_Tablet_Production")

# Define variables
x = m.addVar(lb=30, vtype=GRB.INTEGER, name="laptops")  # Number of laptops produced
y = m.addVar(vtype=GRB.INTEGER, name="tablets")  # Number of tablets produced

# Objective function: Maximize profit
m.setObjective(200*x + 160*y, GRB.MAXIMIZE)

# Constraints
m.addConstr(20*x + 15*y <= 1200, "manufacturing_time")
m.addConstr(3*x + 2*y <= 150, "silicon_availability")

# Solve the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution found: Produce {x.x} laptops and {y.x} tablets.")
else:
    print("No optimal solution found. The model is either infeasible or unbounded.")
```