To solve this problem, we need to:

1. Graph the feasible region defined by the constraints.
2. Identify the vertices of the feasible region.
3. Evaluate the objective function $z = 10x + 3y$ at each vertex to find the minimum and maximum values.

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Steps:

1. Constraints:

1. $x \geq 0$: This restricts the feasible region to the right side of the $y$-axis.
2. $2x + 3y \geq 6$: Rearranged to $y \geq \frac{6 - 2x}{3}$.
3. $3x - y \leq 9$: Rearranged to $y \geq 3x - 9$.
4. $x + 4y \leq 16$: Rearranged to $y \leq \frac{16 - x}{4}$.

2. Feasible Region:

The feasible region is the intersection of all the constraints. We'll identify the vertices by solving the pairwise intersections of these lines.

3. Solve for Intersection Points:

Using the equations of the lines, find the points of intersection:

1. $2x + 3y = 6$ and $3x - y = 9$
2. $2x + 3y = 6$ and $x + 4y = 16$
3. $3x - y = 9$ and $x + 4y = 16$
4. Include boundary conditions $x = 0$ as needed.

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Let's calculate these intersections explicitly.import sympy as sp

Define variables

x, y = sp.symbols('x y')

Define the constraints

constraint1 = 2x + 3y - 6 # 2x + 3y = 6 constraint2 = 3x - y - 9 # 3x - y = 9 constraint3 = x + 4y - 16 # x + 4y = 16

Solve intersections

intersection1 = sp.solve([constraint1, constraint2], (x, y)) intersection2 = sp.solve([constraint1, constraint3], (x, y)) intersection3 = sp.solve([constraint2, constraint3], (x, y))

Include x = 0 boundary

boundary1 = sp.solve([constraint1, x], (x, y)) boundary2 = sp.solve([constraint2, x], (x, y)) boundary3 = sp.solve([constraint3, x], (x, y))

Collect all intersection points

vertices = [ intersection1, intersection2, intersection3, boundary1, boundary2, boundary3 ]

vertices[{x: 3, y: 0}, {x: -24/5, y: 26/5}, {x: 4, y: 3}, {x: 0, y: 2}, {x: 0, y: -9}, {x: 0, y: 4}][{x: 3, y: 0}, {x: -24/5, y: 26/5}, {x: 4, y: 3}, {x: 0, y: 2}, {x: 0, y: -9}, {x: 0, y: 4}]successThe intersection points (vertices) of the feasible region are:

1. $(3, 0)$
2. $\left(-\frac{24}{5}, \frac{26}{5}\right)$ (not valid since $x \geq 0$)
3. $(4, 3)$
4. $(0, 2)$
5. $(0, -9)$ (not valid since $y \geq 0$)
6. $(0, 4)$

The valid vertices are:

• $(3, 0)$
• $(4, 3)$
• $(0, 2)$
• $(0, 4)$

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4. Evaluate the Objective Function

Now evaluate $z = 10x + 3y$ at each valid vertex:

• At $(3, 0)$: $z = 10(3) + 3(0) = 30$
• At $(4, 3)$: $z = 10(4) + 3(3) = 49$
• At $(0, 2)$: $z = 10(0) + 3(2) = 6$
• At $(0, 4)$: $z = 10(0) + 3(4) = 12$

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Results:

• Maximum value: $z = 49$ at $(4, 3)$
• Minimum value: $z = 6$ at $(0, 2)$

Let me know if you'd like a graph or further details. Here are 5 questions to expand on this topic:

1. How do we ensure all constraints are satisfied in the feasible region?
2. What happens if there is no intersection of constraints?
3. How is the feasible region affected by adding or removing constraints?
4. Can the objective function have multiple optimal solutions? When?
5. What changes if we minimize the objective function instead?

Tip: Always check boundary conditions when graphing constraints, especially $x \geq 0$ and $y \geq 0$.