Question

1. given this graph of the feasible region, what is the maximum value of the function $p = 3x + y$? view image description \bigcirc 41 \bigcirc 9 \bigcirc 11 \bigcirc 34

Explanation:

Step1: Identify vertices of feasible region

From the graph, the vertices (corner points) of the feasible region are: Let's assume the coordinates (by grid analysis). Let's find the coordinates:

• First vertex: Let's say (2, 5) [from the left vertical and horizontal lines], wait, no, looking at the grid, the vertical line at x=12? Wait, no, the grid has x from 0 to 20, y from -5 to 5. Wait, the purple region's vertices: let's see the black dots. Let's list the vertices:

1. (2, 5)? Wait, no, maybe (2, 4)? Wait, no, let's check the graph. Wait, the feasible region is a polygon with vertices. Let's assume the vertices are (2, 4), (12, 5), (12, -1)? Wait, no, the y-axis: the top horizontal line is y=5, vertical line at x=12, and the line y = -x + 6? Wait, no, the line equation: let's find the slope. From (2, 4) to (12, -1): slope is (-1 -4)/(12 -2) = -5/10 = -0.5, so equation y -4 = -0.5(x -2) → y = -0.5x +1 +4 → y = -0.5x +5? Wait, no, maybe better to get the actual vertices. Wait, the feasible region's vertices: let's look at the black dots. The top left dot: (2, 5)? Wait, no, the grid: each square is 1 unit. So x=2, y=5; x=12, y=5; x=12, y=-1 (since the line passes through (12, -1) and (2, 4)? Wait, no, the line at the bottom: when x=12, y is -1? Wait, no, the graph shows the feasible region is bounded by y=5 (top), x=12 (right), and the line y = -x + 6? Wait, no, let's calculate the vertices properly.

Wait, the function is P = 3x + y. To find the maximum, we evaluate P at each vertex.

Let's find the vertices:

1. Top left vertex: Let's say (2, 5) (x=2, y=5)
2. Top right vertex: (12, 5) (x=12, y=5)
3. Bottom right vertex: (12, -1) (x=12, y=-1) Wait, no, the line intersects x=12 at y=? Let's find the line equation. The line passes through (2, 4) and (12, -1)? Wait, no, maybe the line is y = -x + 6? Wait, when x=2, y=4: 4 = -2 +6 → 4=4, yes. When x=12, y= -12 +6 = -6? No, that's not. Wait, maybe the line is y = -0.5x + 5. When x=2, y=4: 4 = -1 +5 → 4=4. When x=12, y= -6 +5 = -1. Yes, that works. So the line is y = -0.5x +5.

So the vertices are:

• Intersection of y=5 and x=2: (2, 5)
• Intersection of y=5 and x=12: (12, 5)
• Intersection of x=12 and y = -0.5x +5: x=12, y= -0.5*12 +5 = -6 +5 = -1 → (12, -1)
• Intersection of y = -0.5x +5 and x=2: (2, 4)? Wait, no, x=2, y= -0.5*2 +5 = -1 +5 =4 → (2, 4). Wait, so the feasible region is a quadrilateral? No, maybe a triangle. Wait, the purple region: top is y=5, right is x=12, and the line from (2, 4) to (12, -1), and left is x=2? Wait, no, the graph shows the feasible region is bounded by three lines: y=5 (horizontal), x=12 (vertical), and the line y = -0.5x +5 (slanted). So the vertices are (2, 5), (12, 5), and (12, -1)? No, (2, 5) is on y=5 and x=2, (12, 5) is on y=5 and x=12, and (12, -1) is on x=12 and the slanted line, and (2, 4) is on x=2 and the slanted line. Wait, maybe the feasible region is a polygon with vertices (2, 5), (12, 5), (12, -1), (2, 4). But that's a quadrilateral. But to find the maximum of P=3x + y, we evaluate at each vertex.

Wait, let's check the options: 41, 9, 11, 34.

Evaluate P at (12, 5): P = 3*12 +5 = 36 +5 =41.

At (2, 5): P=3*2 +5=6+5=11.

At (12, -1): P=3*12 + (-1)=36-1=35.

At (2, 4): P=3*2 +4=6+4=10.

So the maximum is 41, which is option A.

Step2: Evaluate P at each vertex

• Vertex (2,5): \( P = 3(2) + 5 = 6 + 5 = 11 \)
• Vertex (12,5): \( P = 3(12) + 5 = 36 + 5 = 41 \)
• Vertex (12,-1): \( P = 3(12) + (-1) = 36 - 1 = 35 \)
• Vertex (2,4): \( P = 3(2) + 4 = 6 + 4 = 10 \)

The largest value is 41.

Answer:

41 (corresponding to the option with 41, e.g., A. 41 if options are labeled, but here the option is 41)