Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.) Maximize p = x + 2y subject to x + 6y ≤ 8 3x + y ≤ 7 x ≥ 0, y ≥ 0. p = (x, y)

We are tasked with solving the following Linear Programming (LP) problem:

Objective function:

Maximize $p = x + 2y$

Subject to constraints:

1. $x + 6y \leq 8$
2. $3x + y \leq 7$
3. $x \geq 0$
4. $y \geq 0$

Step 1: Graph the feasible region

We need to plot the inequalities to identify the feasible region.

1. From $x + 6y \leq 8$:

   • Solve for $y$: $y \leq \frac{8 - x}{6}$
   • This is a line with slope $-\frac{1}{6}$ and intercept $(0, \frac{8}{6} = 1.33)$.

2. From $3x + y \leq 7$:

   • Solve for $y$: $y \leq 7 - 3x$
   • This is a line with slope $-3$ and intercept $(0, 7)$.

3. Non-negativity constraints $x \geq 0$ and $y \geq 0$ imply the solution must be in the first quadrant.

Step 2: Find the corner points

We now find the intersection points of these boundary lines and check the values at each corner.

• Intersection of $x + 6y = 8$ and $3x + y = 7$:

  • Solve the system:

    1. $x + 6y = 8$
    2. $3x + y = 7$

    Substituting $y = 7 - 3x$ into $x + 6y = 8$: $x + 6(7 - 3x) = 8 \implies x + 42 - 18x = 8 \implies -17x = -34 \implies x = 2$ Substituting $x = 2$ back into $3x + y = 7$: $3(2) + y = 7 \implies y = 1$ So, the intersection point is $(2, 1)$.

• Intersection of $x + 6y = 8$ and the $y$-axis ($x = 0$): $0 + 6y = 8 \implies y = \frac{8}{6} = \frac{4}{3}$ So, the point is $(0, \frac{4}{3})$.

• Intersection of $3x + y = 7$ and the $x$-axis ($y = 0$): $3x + 0 = 7 \implies x = \frac{7}{3}$ So, the point is $(\frac{7}{3}, 0)$.

Step 3: Evaluate the objective function at each vertex

Now, we evaluate $p = x + 2y$ at the vertices of the feasible region:

• At $(0, \frac{4}{3})$: $p = 0 + 2\left(\frac{4}{3}\right) = \frac{8}{3} \approx 2.67$
• At $(\frac{7}{3}, 0)$: $p = \frac{7}{3} + 2(0) = \frac{7}{3} \approx 2.33$
• At $(2, 1)$: $p = 2 + 2(1) = 4$

Step 4: Conclusion

The maximum value of $p$ occurs at $(2, 1)$, where $p = 4$.

Thus, the optimal solution is: $p = 4 \text{ at } (x, y) = (2, 1)$

If you have any further questions or would like details, feel free to ask!

Here are 5 related questions:

1. What is the geometric significance of the feasible region in an LP problem?
2. How does the slope of a constraint affect the shape of the feasible region?
3. Why is the optimal solution always at a vertex of the feasible region in LP problems?
4. How can we handle LP problems with more than two variables graphically?
5. What methods are used to solve LP problems when constraints form a complex polyhedron?

Tip: In linear programming, the feasible region is always convex, and the maximum or minimum value of the objective function must occur at a vertex or along an edge if unbounded.