To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. Let's denote the number of dining tables as \(x_1\) and the number of bed frames as \(x_2\).

The objective is to maximize profit. Given that each dining table yields a profit of $300 and each bed frame yields a profit of $400, the objective function can be represented algebraically as:
\[ \text{Maximize:} \quad 300x_1 + 400x_2 \]

The constraints are based on the availability of oak and mahogany wood. Each dining table requires 3 units of oak wood and 5 units of mahogany wood, while each bed frame requires 5 units of oak wood and 2 units of mahogany wood. The total available oak wood is 300 units, and the total available mahogany wood is 400 units. Thus, we have two main constraints:
1. For oak wood: \(3x_1 + 5x_2 \leq 300\)
2. For mahogany wood: \(5x_1 + 2x_2 \leq 400\)

Additionally, since the number of dining tables and bed frames cannot be negative, we have:
\[ x_1 \geq 0, \quad x_2 \geq 0 \]

Therefore, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'dining tables'), ('x2', 'bed frames')],
    'objective_function': '300*x1 + 400*x2',
    'constraints': ['3*x1 + 5*x2 <= 300', '5*x1 + 2*x2 <= 400', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we will use the following code:
```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name="dining_tables", vtype=GRB.CONTINUOUS, lb=0)
x2 = m.addVar(name="bed_frames", vtype=GRB.CONTINUOUS, lb=0)

# Set the objective function
m.setObjective(300*x1 + 400*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(3*x1 + 5*x2 <= 300, name="oak_wood_constraint")
m.addConstr(5*x1 + 2*x2 <= 400, name="mahogany_wood_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Dining Tables: {x1.x}")
    print(f"Bed Frames: {x2.x}")
    print(f"Maximum Profit: {m.objVal}")
else:
    print("No optimal solution found")
```