Graph the system of constraints and find the value of x and y that maximize the objective function.
[*20
v20
Constraints „s!
-2+2
5
52y+x
Objective function: C = 7x - 3y
(1 point)
• (2.5. 2.5)
• (0, 2)
• (0, 0)
• (5, 0)
To graph the system of constraints, we need to plot the equations and inequalities.
First, let's plot the equation -2x + 2y = 20.
To do this, we can find two points on the line and draw a straight line passing through them.
When x = 0, we have -2(0) + 2y = 20, which gives us y = 10. So one point on the line is (0, 10).
When y = 0, we have -2x + 2(0) = 20, which gives us x = -10. So another point on the line is (-10, 0).
Plotting these points and drawing a line passing through them, we get:
Next, let's plot the inequality 5x + 2y <= 20.
To do this, we find two points on the line and draw a dashed line passing through them (since it's an inequality).
When x = 0, we have 5(0) + 2y <= 20, which gives us 2y <= 20. Dividing both sides by 2, we get y <= 10. So one point on the line is (0, 10).
When y = 0, we have 5x + 2(0) <= 20, which gives us 5x <= 20. Dividing both sides by 5, we get x <= 4. So another point on the line is (4, 0).
Plotting these points and drawing a dashed line passing through them, we get:
The shaded region represents the area where both constraints are satisfied.
Finally, let's find the value of x and y that maximize the objective function C = 7x - 3y within this feasible region.
To do this, we need to find the intersection of the lines/lines and inequalities. Looking at the graph, we see that the feasible region is a triangle with vertices at (0, 0), (0, 10), and (4, 0).
We evaluate the objective function C = 7x - 3y at each corner of the triangle:
C(0, 0) = 7(0) - 3(0) = 0
C(0, 10) = 7(0) - 3(10) = -30
C(4, 0) = 7(4) - 3(0) = 28
Therefore, the value of x and y that maximize the objective function is (4, 0).
To graph the system of constraints, we need to convert the inequalities into equations and plot them on a graph.
The given constraints are:
Constraint 1: -2x + 2y ≤ 20
Constraint 2: 5x + 2y ≤ 20
Let's solve each constraint for y:
Constraint 1: -2x + 2y ≤ 20
Adding 2x to both sides: 2y ≤ 2x + 20
Dividing by 2: y ≤ x + 10
Constraint 2: 5x + 2y ≤ 20
Subtracting 5x from both sides: 2y ≤ -5x + 20
Dividing by 2: y ≤ -2.5x + 10
Now, let's plot these inequalities on a graph:
First, draw the line y = x + 10. This is a straight line with a y-intercept of 10 and a slope of 1.
Next, draw the line y = -2.5x + 10. This is a straight line with a y-intercept of 10 and a slope of -2.5.
Shade the region below both lines, as the inequalities are less than or equal to.
The feasible region will be the intersection of the shaded region and the first quadrant (x ≥ 0, y ≥ 0).
To find the value of x and y that maximize the objective function C = 7x - 3y, we need to evaluate the function at the vertices of the feasible region.
The possible vertices are:
A: (2.5, 2.5)
B: (0, 2)
C: (0, 0)
D: (5, 0)
Evaluate the objective function at each vertex:
C(A) = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10
C(B) = 7(0) - 3(2) = 0 - 6 = -6
C(C) = 7(0) - 3(0) = 0 - 0 = 0
C(D) = 7(5) - 3(0) = 35 - 0 = 35
Out of these values, the maximum value of the objective function is 35, which occurs at vertex D: (5, 0).
Therefore, the value of x and y that maximize the objective function are x = 5 and y = 0.
To graph the system of constraints, we need to convert the constraints into inequality equations and plot them on a coordinate plane.
1. The first constraint is -2x + 2y ≤ 20. We can rewrite this as 2y ≤ 2x + 20 and then divide by 2 to get y ≤ x + 10. The inequality equation is y ≤ x + 10.
2. The second constraint is 5x + 2y ≤ 20. We can rewrite this as 2y ≤ -5x + 20 and then divide by 2 to get y ≤ -2.5x + 10. The inequality equation is y ≤ -2.5x + 10.
3. The third constraint is 5x + 2y ≥ 5. We can rewrite this as 2y ≥ -5x + 5 and then divide by 2 to get y ≥ -2.5x + 2.5. The inequality equation is y ≥ -2.5x + 2.5.
Now let's plot these inequalities on a graph:
First, draw a line for y = x + 10, and shade the region below the line.
Second, draw a line for y = -2.5x + 10, and shade the region below the line.
Third, draw a line for y = -2.5x + 2.5, and shade the region above the line.
The shaded region where all three regions overlap represents the feasible region.
To find the values of x and y that maximize the objective function C = 7x - 3y, evaluate the objective function at the vertices of the feasible region.
Now let's find the vertices and evaluate the objective function:
1. (2.5, 2.5):
C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10
2. (0, 2):
C = 7(0) - 3(2) = -6
3. (0, 0):
C = 7(0) - 3(0) = 0
4. (5, 0):
C = 7(5) - 3(0) = 35
It appears that the pair (2.5, 2.5) maximizes the objective function C = 7x - 3y among the given options.
Your computer-supply store sells two types of inkjet printers. The first, type A, costs $237 and you make a $22 profit on each one.
The second, type B, costs $122 and you make a $19 profit on each one. You can order no more than 120 printers this month, and you need to make at least $2,400 profit on them. If you must order at least one of each type of printer, how many of each type of printer should you order if you want to minimize your cost?
(1 point)
• 69 of type A : 51 of type B
• 40 of type A : 80 of type B
• 51 of type A : 69 of type B
• 80 of type A : 40 of type B
Let's assume the number of type A printers you order is represented by the variable "x", and the number of type B printers you order is represented by the variable "y".
According to the given information, we can form the following equations and inequalities:
1. Cost constraint: 237x + 122y ≤ 2400 (total cost should be less than or equal to $2400)
2. Quantity constraint: x + y ≤ 120 (total number of printers should be less than or equal to 120)
3. Profit constraint: 22x + 19y ≥ 2400 (total profit should be greater than or equal to $2400)
4. Non-negativity constraint: x ≥ 0, y ≥ 0 (the number of printers cannot be negative)
To minimize the cost, we need to solve this system of equations and inequalities.
Let's graph the feasible region for these constraints:
For the inequality 237x + 122y ≤ 2400, we can rewrite it as y ≤ (-237/122)x + 19.67.
For the inequality x + y ≤ 120, we can rewrite it as y ≤ -x + 120.
For the inequality 22x + 19y ≥ 2400, we can rewrite it as y ≥ (-22/19)x + 126.32.
Plotting these lines on a graph and shading the region that satisfies all constraints, we get:
Now, we need to evaluate the corner points of the feasible region to find the minimum cost.
The corner points are:
(0, 0) -> Cost: 237 * 0 + 122 * 0 = 0
(0, 109) -> Cost: 237 * 0 + 122 * 109 = 13318
(40, 80) -> Cost: 237 * 40 + 122 * 80 = 21440
(69, 51) -> Cost: 237 * 69 + 122 * 51 = 20655
(120, 0) -> Cost: 237 * 120 + 122 * 0 = 28440
The minimum cost occurs when you order 40 of type A and 80 of type B printers. Therefore, the answer is:
40 of type A : 80 of type B.
A factory can produce two products, × and y, with a profit approximated by P = 14 + 22y - 900. The production of y can exceed x by no more than 200 units. Moreover, production levels are limited by the formula × + 2y ≤ 1600. What production levels yield maximum profit?
(1 point)
select one answer no explanations or essays
• x = 400; y = 600
• x = 0; y = 0
• x = 1,600; y = 0
• × = 0; y = 200
• x = 400; y = 600
wrong
Apologies for the incorrect response. The correct answer is:
• x = 800; y = 400