We need to minimize $f(x_1, x_2) = 5x_1 + x_2$ subject to the given constraints. Let’s carefully rewrite the constraints and analyze the problem.

Problem Formulation

Constraints:

1. $3x_1 + 4x_2 \geq 13$
2. $4x_1 + x_2 \geq 5$

Step 1: Convert the constraints to equalities for corner points

To find the feasible region, we solve the system of linear inequalities. The feasible region's corner points are intersections of the constraint lines and the axes.

Equation 1: $3x_1 + 4x_2 = 13$

• When $x_1 = 0$: $4x_2 = 13 \implies x_2 = \frac{13}{4}$.
• When $x_2 = 0$: $3x_1 = 13 \implies x_1 = \frac{13}{3}$.

So, the line intersects at points $(0, \frac{13}{4})$ and $(\frac{13}{3}, 0)$.

Equation 2: $4x_1 + x_2 = 5$

• When $x_1 = 0$: $x_2 = 5$.
• When $x_2 = 0$: $4x_1 = 5 \implies x_1 = \frac{5}{4}$.

So, the line intersects at points $(0, 5)$ and $(\frac{5}{4}, 0)$.

Step 2: Find the intersection of the two lines

Solve the system: $3x_1 + 4x_2 = 13$ $4x_1 + x_2 = 5$

From the second equation, $x_2 = 5 - 4x_1$. Substitute this into the first equation: $3x_1 + 4(5 - 4x_1) = 13$ $3x_1 + 20 - 16x_1 = 13$ $-13x_1 + 20 = 13 \implies -13x_1 = -7 \implies x_1 = \frac{7}{13}$ Substitute $x_1 = \frac{7}{13}$ into $x_2 = 5 - 4x_1$: $x_2 = 5 - 4\left(\frac{7}{13}\right) = 5 - \frac{28}{13} = \frac{65}{13} - \frac{28}{13} = \frac{37}{13}.$

So, the intersection point is: $\left( \frac{7}{13}, \frac{37}{13} \right).$

Step 3: Evaluate $f(x_1, x_2)$ at each vertex of the feasible region

Vertices of the feasible region are:

1. $(0, \frac{13}{4})$
2. $(\frac{13}{3}, 0)$
3. $(0, 5)$
4. $(\frac{5}{4}, 0)$
5. $\left( \frac{7}{13}, \frac{37}{13} \right)$

Vertex 1: $(0, \frac{13}{4})$

$f(0, \frac{13}{4}) = 5(0) + \frac{13}{4} = \frac{13}{4}.$

Vertex 2: $(\frac{13}{3}, 0)$

$f(\frac{13}{3}, 0) = 5\left(\frac{13}{3}\right) + 0 = \frac{65}{3}.$

Vertex 3: $(0, 5)$

$f(0, 5) = 5(0) + 5 = 5.$

Vertex 4: $(\frac{5}{4}, 0)$

$f(\frac{5}{4}, 0) = 5\left(\frac{5}{4}\right) + 0 = \frac{25}{4}.$

Vertex 5: $\left( \frac{7}{13}, \frac{37}{13} \right)$

$f\left(\frac{7}{13}, \frac{37}{13}\right) = 5\left(\frac{7}{13}\right) + \frac{37}{13} = \frac{35}{13} + \frac{37}{13} = \frac{72}{13}.$

Step 4: Find the minimum value

The values of $f(x_1, x_2)$ at the vertices are:

• $\frac{13}{4}$
• $\frac{65}{3}$
• $5$
• $\frac{25}{4}$
• $\frac{72}{13}$

The smallest value is $5$, which occurs at $(0, 5)$.

Final Answer:

The minimum value is: $\boxed{5}.$

Do you have any questions or would you like further clarification? Here are related questions for further practice:

1. How do you identify the feasible region in a linear programming problem?
2. What is the graphical interpretation of the intersection point of two constraints?
3. How do you determine if a feasible region is bounded or unbounded?
4. Can you describe the role of corner points in finding the optimal solution?
5. What changes if one of the constraints is a strict inequality?

Tip: Always check if the feasible region is bounded before concluding whether a minimum or maximum exists.