\( 1 \leftarrow \quad \begin{array}{l}\text { A set of constraints and a profit function are given. Answer } \\ \text { parts (a) and (b). } \\ 3 x+4 y \geq 20 \\ 6 x+3 y \leq 30 \\ x \geq 2 \\ y \geq 2 \\ P=2.67 x+1.32 y \\ \text { a) Graph the constraints, shade the feasible region, and } \\ \text { determine its vertices. }\end{array} \)

We are given the system of constraints   1) 3x + 4y ≥ 20   2) 6x + 3y ≤ 30   3) x ≥ 2   4) y ≥ 2 with the profit function P = 2.67x + 1.32y. In part (a) we need to graph these constraints, shade the feasible region, and determine its vertices. Let’s go step by step. ──────────────────────────── Step 1. Rewrite the constraint boundaries as equalities: • Boundary of (1): 3x + 4y = 20   Solve for y: y = (20 – 3x)/4 • Boundary of (2): 6x + 3y = 30   Solve for y: y = (30 – 6x)/3 = 10 – 2x • Boundaries (3) and (4) are the vertical line x = 2 and the horizontal line y = 2, respectively. ──────────────────────────── Step 2. Determine where the “sides” of the feasible region lie: • Constraint (1) is “≥”, so the feasible region lies on or above the line y = (20 – 3x)/4. • Constraint (2) is “≤”, so the region lies on or below the line y = 10 – 2x. • Constraints (3) and (4) require that we are to the right of x = 2 and above y = 2. ──────────────────────────── Step 3. Find the Intersection Points (Vertices) a) Intersection of 3x + 4y = 20 and 6x + 3y = 30:   Solve the system:    3x + 4y = 20     (Equation A)    6x + 3y = 30     (Equation B) Multiply Equation A by 2:   6x + 8y = 40     (Equation A′) Subtract Equation B from Equation A′:   (6x + 8y) – (6x + 3y) = 40 – 30   5y = 10  ⇒ y = 2 Substitute y = 2 into Equation A:   3x + 4(2) = 20  ⇒ 3x + 8 = 20  ⇒ 3x = 12  ⇒ x = 4 Thus, one vertex is (4, 2). b) Intersection of 3x + 4y = 20 and x = 2:   Substitute x = 2 into 3x + 4y = 20:    3(2) + 4y = 20  ⇒ 6 + 4y = 20  ⇒ 4y = 14  ⇒ y = 3.5 This gives the vertex (2, 3.5). c) Intersection of 6x + 3y = 30 and x = 2:   Substitute x = 2 into 6x + 3y = 30:    6(2) + 3y = 30  ⇒ 12 + 3y = 30  ⇒ 3y = 18  ⇒ y = 6 This gives the vertex (2, 6). ──────────────────────────── Step 4. Check the Constraint y = 2 Although we already have the line y = 2 as a boundary, its intersection with other constraints has given us (4, 2) (which we already found as the intersection of the two non-vertical/horizontal lines) so no additional vertex is needed here. ──────────────────────────── Step 5. Sketching the Graph 1. Draw the line 3x + 4y = 20.   • For x=0, 4y = 20 ⇒ y = 5.   • For y=0, 3x = 20 ⇒ x ≈ 6.67.   Shade above this line (since 3x + 4y should be ≥ 20). 2. Draw the line 6x + 3y = 30.   • For x=0, 3y = 30 ⇒ y = 10.   • For y=0, 6x = 30 ⇒ x = 5.   Shade below this line (since 6x + 3y should be ≤ 30). 3. Draw the vertical line x = 2 and the horizontal line y = 2.   Shade to the right of x = 2 and above y = 2. The feasible region is the area that simultaneously satisfies all these conditions. When plotted, you will see that the region is a triangle with vertices at   (2, 6), (2, 3.5), and (4, 2). ──────────────────────────── Final Answer for Part (a): The feasible region is the triangular region bounded by the lines   x = 2, y = (20 – 3x)/4, and y = 10 – 2x. Its vertices are:   (2, 6), (2, 3.5), and (4, 2). These are the points you would plot and label when drawing the graph.