To solve the linear programming problem and find the corner points, we will follow these steps:

Problem Statement:

Maximize:
$C = 20x_1 + 30x_2$

Subject to the constraints:

1. $6x_1 + x_2 \leq 12$
2. $4x_1 + 4x_2 \leq 49$
3. $7x_1 + 7x_2 \leq 33$

And the non-negativity constraints:

• $x_1 \geq 0$
• $x_2 \geq 0$

Steps to Find Corner Points:

1. Convert Inequalities to Equalities:
   The corner points occur where the boundary lines of the inequalities intersect. So, we will treat the inequalities as equalities to form systems of linear equations.

2. Solve Systems of Equations:
   Find the points where the lines intersect by solving pairs of equations.

3. Check for Feasibility:
   Once the intersection points are found, ensure they satisfy the original inequalities.

1. Intersection of $6x_1 + x_2 = 12$ and $4x_1 + 4x_2 = 49$

Solve the system: 6x_1 + x_2 = 12 \tag{1} 4x_1 + 4x_2 = 49 \tag{2}

From equation (2), divide both sides by 4: x_1 + x_2 = 12.25 \tag{3}

Now, subtract equation (1) from equation (3): $(x_1 + x_2) - (6x_1 + x_2) = 12.25 - 12$ $-5x_1 = 0.25$ $x_1 = -0.05 \quad (\text{not feasible since } x_1 \geq 0)$ Thus, no valid intersection here.

2. Intersection of $6x_1 + x_2 = 12$ and $7x_1 + 7x_2 = 33$

Solve the system: 6x_1 + x_2 = 12 \tag{4} 7x_1 + 7x_2 = 33 \tag{5}

Divide equation (5) by 7: x_1 + x_2 = 4.71 \tag{6}

Now, subtract equation (4) from equation (6): $(x_1 + x_2) - (6x_1 + x_2) = 4.71 - 12$ $-5x_1 = -7.29$ $x_1 = 1.458$

Substitute $x_1 = 1.458$ into equation (4): $6(1.458) + x_2 = 12$ $8.748 + x_2 = 12$ $x_2 = 3.252$

So, the point of intersection is $(1.458, 3.252)$.

3. Intersection of $4x_1 + 4x_2 = 49$ and $7x_1 + 7x_2 = 33$

Solve the system: 4x_1 + 4x_2 = 49 \tag{7} 7x_1 + 7x_2 = 33 \tag{8}

Divide both equations by their respective coefficients: x_1 + x_2 = 12.25 \tag{9} x_1 + x_2 = 4.71 \tag{10}

There is no solution for this system as both equations contradict each other. So, there is no intersection between these two lines.

4. Intersection with the axes:

Check for points where $x_1 = 0$ or $x_2 = 0$.

For $x_1 = 0$:

• From $6x_1 + x_2 = 12$, $x_2 = 12$.
• From $4x_1 + 4x_2 = 49$, $4x_2 = 49 \Rightarrow x_2 = 12.25$, which is not feasible (as $x_2 \leq 12$).
• From $7x_1 + 7x_2 = 33$, $7x_2 = 33 \Rightarrow x_2 = 4.71$.

So, valid point: $(0, 4.71)$.

For $x_2 = 0$:

• From $6x_1 + x_2 = 12$, $6x_1 = 12 \Rightarrow x_1 = 2$.
• From $4x_1 + 4x_2 = 49$, $4x_1 = 49 \Rightarrow x_1 = 12.25$, which is not feasible.
• From $7x_1 + 7x_2 = 33$, $7x_1 = 33 \Rightarrow x_1 = 4.71$.

So, valid point: $(2, 0)$.

Conclusion:

The feasible corner points are:

1. $(0, 4.71)$
2. $(2, 0)$
3. $(1.458, 3.252)$

These are the corner points of the feasible region.

Would you like to go over the calculations in more detail or explore further steps? Here are some related questions:

1. How do we test which corner point gives the maximum value for $C$?
2. How can we handle linear programming problems with multiple variables?
3. What other methods are available to solve linear programming problems besides the graphical method?
4. How do we determine if a solution is optimal in a real-world scenario?
5. What happens if the constraints are changed or modified?

Tip: Always check the feasibility of each corner point against all constraints to ensure it lies within the feasible region.