Multiple Choice
Graph the system of constraints and find the value of x and y that maximize the objective function.
* ≥ 0
y ≥0
Constraints us!
§*+2
52 y+ x
Objective Function: C = 1x - 3y
(1 point)
• (0, 0)
• (2, 3)
• (5, 0)
• (0, 3)
To graph the system of constraints, we need to plot the lines representing the equations and shade the feasible region.
The first constraint is y ≥ 0, which is a horizontal line passing through the y-axis at y = 0. We shade the region above this line.
The second constraint is 52y + x ≤ 2, which can be rewritten as x ≤ -52y + 2. To graph this line, we plot two points and draw a line through them. Let's choose y = 0 and y = 1 as our points:
If y = 0, x = -52(0) + 2 = 2, so the point (2, 0) is on the line.
If y = 1, x = -52(1) + 2 = -50, so the point (-50, 1) is on the line.
We draw a line passing through these two points and shade the region below this line.
Now, we need to find the intersection of these two shaded regions, which will be our feasible region.
Looking at the graph, the feasible region consists of the points (0, 0), (2, 0), and (0, 3).
To find the value of x and y that maximize the objective function, we can substitute the coordinates of each point into the objective function and see which one gives the highest value of C.
For (0, 0):
C = 1(0) - 3(0) = 0
For (2, 0):
C = 1(2) - 3(0) = 2
For (0, 3):
C = 1(0) - 3(3) = -9
Therefore, the pair of values that maximize the objective function is (2, 0).
Answer: (2, 0)
To graph the system of constraints, we will first plot the lines representing the inequalities.
The constraint "y ≥ 0" represents the region above the x-axis.
The constraint "x + 2y ≥ 0" can be rewritten as "y ≥ -x/2," which represents the region below the line with a slope of -1/2 passing through the origin.
The constraints together form a region in the first quadrant.
Now, let's find the feasible region by finding the intersection of the two lines.
Setting y = -x/2 in the inequality y ≥ -x/2, we have: -x/2 ≥ 0
Multiplying both sides by -2 (and flipping the inequality sign), we get: x ≤ 0
So, the feasible region lies in the region x ≤ 0 and y ≥ 0.
To find the value of x and y that maximize the objective function C = 1x - 3y, we need to find the highest point in the feasible region. Since both x and y are non-negative, the highest point in the feasible region will be at the intersection of the lines x = 0 and y = 0.
Therefore, the value of x and y that maximize the objective function is (0, 0).
Answer: (0, 0)
To find the value of x and y that maximize the objective function, we first need to graph the system of constraints and then identify the feasible region. The objective function will be maximized at a point within this feasible region.
Let's graph the system of constraints:
1) We have the constraint: * ≥ 0. This means that x can take on any non-negative value. We will draw a vertical line to represent this constraint.
2) We also have the constraint: y ≥ 0. This means that y can take on any non-negative value. We will draw a horizontal line to represent this constraint.
3) The constraint §*+2 represents a linear inequality. To graph it, we'll rewrite it in slope-intercept form: y ≤ -2x + 5. We'll graph the line y = -2x + 5 and shade the region below the line to represent the inequality.
Now, we need to find the feasible region, which is the region where all the constraints overlap. The feasible region is the shaded region where the vertical line, horizontal line, and the shaded region below the line y = -2x + 5 intersect.
Next, we can evaluate the objective function C = 1x - 3y at the corner points of the feasible region to find the maximum value. We substitute the x and y values of each corner point into the objective function and choose the maximum value.
Let's evaluate the objective function C = 1x - 3y at the corner points of the feasible region:
1) (0, 0): C = 1(0) - 3(0) = 0
2) (2, 3): C = 1(2) - 3(3) = -7
3) (5, 0): C = 1(5) - 3(0) = 5
4) (0, 3): C = 1(0) - 3(3) = -9
As we can see, the maximum value of the objective function is 5 and it occurs at the point (5, 0). Therefore, the correct answer is option (c) (5, 0).