Question

a manufacturing machine has two processes. one of them is repeated 4 times and the second only once. the entire cycle can take no longer than 3 minutes. which graph represents the overall equation represented by this scenario? (all points may not apply to the scenario.)

Explanation:

Step1: Define Variables

Let \( x \) be the time for the first process (repeated 4 times) and \( y \) be the time for the second process (repeated once). The total time constraint is \( 4x + y \leq 3 \), which can be rewritten as \( y \leq -4x + 3 \).

Step2: Analyze the Inequality

The boundary line is \( y = -4x + 3 \), with a slope of \(-4\) and a y - intercept of \( 3 \). Since the inequality is \( y \leq -4x + 3 \), we shade the region below the line. Also, since time cannot be negative (\( x\geq0,y\geq0 \)), we consider the first quadrant.

Step3: Analyze the Graphs

- The line \( y=-4x + 3 \) has a steep negative slope. When \( x = 0 \), \( y=3 \); when \( y = 0 \), \( x=\frac{3}{4}=0.75 \).
- We need to check the shading: the region should be where \( y\leq - 4x + 3 \) and \( x\geq0,y\geq0 \). The correct graph should have the line \( y = - 4x+3 \) (with a steep slope) and shading below the line in the first - quadrant - like region (since \( x\) and \( y\) represent time, non - negative). Looking at the options, the graph with the steep line \( y=-4x + 3 \) and shading below (in the region where \( x\geq0,y\geq0 \)) is the one where the shaded area is in the lower part relative to the line and in the first - quadrant - adjacent region. The graph with the red line (steep slope) and shading below the line (consistent with \( y\leq - 4x + 3 \)) and considering non - negative time values is the fourth graph (the one with the shaded area on the left - lower side relative to the line, considering \( x\geq0,y\geq0 \)). Wait, actually, let's re - evaluate. Wait, the original inequality is \( 4x + y\leq3 \), so \( x\) and \( y\) are non - negative (time can't be negative). So the feasible region is in the first quadrant, below the line \( y=-4x + 3 \). The line \( y=-4x + 3 \) has a very steep negative slope. When we look at the graphs, the correct graph should have the line with slope - 4, y - intercept 3, and shading below the line in the first quadrant (where \( x\geq0,y\geq0 \)). Among the given graphs, the graph with the red line (steep slope) and shading in the region that is below the line and in the first - quadrant - like area (since \( x\) and \( y\) are non - negative) is the one where the shaded area is on the side where \( x\geq0,y\geq0 \) and below the line. Let's check the intercepts: when \( x = 0 \), \( y = 3 \); when \( y=0 \), \( x = 3/4=0.75 \). So the line should cross the y - axis at 3 and the x - axis at 0.75. The graph that has this line and shading below (for \( y\leq - 4x + 3 \)) and in the first quadrant (since \( x,y\geq0 \)) is the fourth graph (the one with the shaded area on the left - hand side relative to the line, considering the non - negative time). Wait, actually, let's look at the slope. The slope of \( y=-4x + 3 \) is - 4, which is a very steep negative slope. So the line should be very steep. Among the four graphs, the one with the steepest line (since slope - 4 is very steep) and shading below the line (because \( y\leq - 4x + 3 \)) and in the region where \( x\geq0,y\geq0 \) (so the shaded area is in the lower - left part relative to the line, considering \( x\) and \( y\) non - negative) is the fourth graph (the one with the shaded area on the left side, with the line having a steep slope, y - intercept 3, and x - intercept around 0.75). Wait, maybe I made a mistake earlier. Let's re - express the inequality. The total time is \( 4x + y\leq3 \), so \( y\leq - 4x+3 \). The boundary line is \( y = - 4x + 3 \). The slope is - 4, which is a very steep negative slope. So the line should be going down very steeply from (0,3) to (0.75,0). The shading is below the line. So the correct graph is the one where the line is \( y=-4x + 3 \) (steep slope) and the shading is below the line, in the region where \( x\geq0 \) and \( y\geq0 \). Looking at the options, the fourth graph (the one with the shaded area on the left - lower side, with the red line having a steep slope) is the correct one. Wait, actually, let's check the first - quadrant constraint. Since \( x\) and \( y\) are times, they must be non - negative. So the feasible region is in the first quadrant ( \( x\geq0,y\geq0 \)) and below the line \( y=-4x + 3 \). So the graph should have the line \( y=-4x + 3 \) (steep slope) and shading in the region where \( x\geq0,y\geq0 \) and \( y\leq - 4x + 3 \). The correct graph is the one with the red line (steep slope) and shading below the line, in the area that is in the first - quadrant - adjacent region (since \( x\) and \( y\) are non - negative). So the answer is the fourth graph (the one with the shaded area on the left side, with the red line of steep slope).

Answer:

The fourth graph (the one with the shaded area on the left - hand side relative to the steep red line, representing \( y\leq - 4x + 3 \) with \( x\geq0,y\geq0 \))