This is a linear programming problem where you need to maximize $p = x + 2y$ subject to the following constraints:

$x + 6y \leq 20, \quad 6x + y \leq 15, \quad x \geq 0, \quad y \geq 0.$

Steps to Solve:

1. Graph the Constraints:

   • $x + 6y = 20$: This line intersects the $x$-axis at $(20, 0)$ and the $y$-axis at $(0, \frac{20}{6} \approx 3.33)$.
   • $6x + y = 15$: This line intersects the $x$-axis at $(2.5, 0)$ and the $y$-axis at $(0, 15)$.
   • $x \geq 0, y \geq 0$: Restrict the feasible region to the first quadrant.

2. Find Corner Points: The feasible region is bounded by the intersection of these lines and the $x$- and $y$-axes:

   • Solve the system of equations:
     • $x + 6y = 20$ and $6x + y = 15$ to find the intersection point.
   • Intersections with axes and each other form the vertices of the feasible region.

3. Evaluate $p$ at Each Vertex: Substitute the coordinates of each corner point into $p = x + 2y$.

Let me calculate the exact corner points and the optimal value. sympy import symbols, Eq, solve

Define variables

x, y = symbols('x y', nonnegative=True)

Define constraints as equations

eq1 = Eq(x + 6y, 20) # x + 6y = 20 eq2 = Eq(6x + y, 15) # 6x + y = 15

Intersection of the two lines

intersection = solve((eq1, eq2), (x, y))

Intersection points with axes

x_intercept_eq1 = solve(eq1.subs(y, 0), x) # x-intercept of eq1 y_intercept_eq1 = solve(eq1.subs(x, 0), y) # y-intercept of eq1 x_intercept_eq2 = solve(eq2.subs(y, 0), x) # x-intercept of eq2 y_intercept_eq2 = solve(eq2.subs(x, 0), y) # y-intercept of eq2

Collect vertices of feasible region

vertices = [ (x_intercept_eq1[0], 0), # x-intercept of eq1 (0, y_intercept_eq1[0]), # y-intercept of eq1 (x_intercept_eq2[0], 0), # x-intercept of eq2 (0, y_intercept_eq2[0]), # y-intercept of eq2 (intersection[x], intersection[y]) # intersection of eq1 and eq2 ]

Evaluate p = x + 2y at each vertex

p_values = [(v[0], v[1], v[0] + 2*v[1]) for v in vertices]

vertices, p_valuesThe vertices of the feasible region and the corresponding values of $p = x + 2y$ are as follows:

1. $(20, 0)$: $p = 20$
2. $\left(0, \frac{10}{3}\right)$: $p = \frac{20}{3} \approx 6.67$
3. $\left(\frac{5}{2}, 0\right)$: $p = \frac{5}{2} = 2.5$
4. $(0, 15)$: $p = 30$ (maximum)
5. $(2, 3)$: $p = 8$

Optimal Solution:

• The maximum value of $p$ is 30, occurring at $(x, y) = (0, 15)$.

Would you like a graph of this feasible region or more explanation?

Related Questions:

1. How are the corner points of the feasible region calculated geometrically?
2. Can this problem be solved using matrix methods (simplex method)?
3. What happens if we change the objective function to $p = 2x + y$?
4. How does the feasibility of the solution change if one constraint becomes an equality?
5. How do you interpret the solution in real-world optimization problems?

Tip:

Always check all vertices of the feasible region when solving linear programming problems, as the optimum is guaranteed to occur at one of them.