To solve this problem, we need to minimize the function $f(x_1, x_2) = 2x_1 + 5x_2$ under the given constraints:

1. $3x_1 + x_2 \geq 10$
2. $x_1 + x_2 \geq 5$

Step 1: Set up the problem

We are dealing with a linear programming (LP) problem where we want to find the minimum value of a linear function subject to linear inequality constraints.

Step 2: Identify feasible region

The constraints define a feasible region in the $x_1$-$x_2$ plane, and we need to find the intersection points of the constraint lines to determine the vertices of this region. Since this is a minimization problem for a linear function, the minimum will be at one of the vertices (corner points) of the feasible region.

1. Equation of constraint 1: $3x_1 + x_2 = 10$
2. Equation of constraint 2: $x_1 + x_2 = 5$

Step 3: Find the intersection points

We solve these equations to find the vertices of the feasible region.

Intersection of constraints 1 and 2:

1. From $x_1 + x_2 = 5$, solve for $x_2 = 5 - x_1$.
2. Substitute into $3x_1 + x_2 = 10$: $3x_1 + (5 - x_1) = 10$ $2x_1 = 5 \Rightarrow x_1 = 2.5$
3. Substitute $x_1 = 2.5$ back into $x_1 + x_2 = 5$: $x_2 = 5 - 2.5 = 2.5$ So, one vertex is $(2.5, 2.5)$.

Points where constraints intersect axes

1. For $3x_1 + x_2 = 10$:

   • Set $x_1 = 0$: $x_2 = 10$ gives the point $(0, 10)$.
   • Set $x_2 = 0$: $x_1 = \frac{10}{3} \approx 3.33$, giving the point $\left(\frac{10}{3}, 0\right)$.

2. For $x_1 + x_2 = 5$:

   • Set $x_1 = 0$: $x_2 = 5$ gives the point $(0, 5)$.
   • Set $x_2 = 0$: $x_1 = 5$ gives the point $(5, 0)$.

Step 4: Evaluate the objective function at each vertex

Now, we calculate $f(x_1, x_2) = 2x_1 + 5x_2$ at each of the points that are feasible under both constraints.

1. $(2.5, 2.5)$: $f(2.5, 2.5) = 2(2.5) + 5(2.5) = 5 + 12.5 = 17.5$

2. $(0, 10)$: Not feasible since $x_1 + x_2 = 10$, which does not satisfy $x_1 + x_2 \geq 5$.

3. $\left(\frac{10}{3}, 0\right)$: $f\left(\frac{10}{3}, 0\right) = 2 \cdot \frac{10}{3} + 5 \cdot 0 = \frac{20}{3} \approx 6.67$

4. $(0, 5)$: $f(0, 5) = 2 \cdot 0 + 5 \cdot 5 = 25$

5. $(5, 0)$: Not feasible as it does not satisfy $3x_1 + x_2 \geq 10$.

Step 5: Identify the minimum

The minimum value of $f(x_1, x_2)$ among the feasible points is at $\left(\frac{10}{3}, 0\right)$, with a value of approximately $6.67$.

Final Answer

The values of $x_1$ and $x_2$ that minimize the function are: $x_1 = \frac{10}{3}, \quad x_2 = 0$ with a minimum value of $f\left(\frac{10}{3}, 0\right) = \frac{20}{3}$.

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Related Questions

1. How would the solution change if we maximized instead of minimized $f(x_1, x_2)$?
2. What if the constraint $3x_1 + x_2 \geq 10$ was instead $3x_1 + x_2 \leq 10$?
3. How do we approach a similar problem if $f(x_1, x_2)$ was nonlinear?
4. Could the feasible region be unbounded, and how would that affect the solution?
5. How does changing the coefficients in $f(x_1, x_2)$ affect the optimal solution?

Tip

In linear programming problems, the optimal value for the objective function always occurs at a vertex of the feasible region. This reduces the need to check all points in the region.