I really don't understand these.. can someone show me the correct way of doing them? (if it helps I will pay over paypal for detailed help or will pay in my services (I am a graphic designer, video editor, and web designer)
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1. 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?
Graph the system of constraints and find the value of x and y that maximize the objective function.
Constraints:
x ≥ 0
y ≥ 0
y ≤ (1)/(5) x + 2
5 ≤ y + x
Objective Function: C = 7x - 3y
a.(2.5, 2.5)
b.(0,2)
c.(0,0)
d.(5,0)
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2. 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?
a. 69 of type A: 51 of type B
b. 40 of type A: 80 of type B
c. 51 of type A: 69 of type B
d. 80 of type A: 40 of type B
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3. A factory can produce two products, x and y, with a profit approximated by P = 14x + 22y – 900. The production of y can exceed x by no more than 200 units. Moreover, production levels are limited by the formula x + 2y ≤ 1600. What production levels yield maximum profit?
a. x = 400; y = 600
b. x = 0; y = 0
c. x = 1,600; y = 0
d. x = 0; y = 200
take a visit to
http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html
and you can play around with this stuff.
Thanks :D I'm 90% sure number 1 is D, but do a good way how I could do 2 or 3? If you don't mind explaining to me?
I'd be happy to help you understand these problems. Let's go through each of them step by step.
1. To solve this problem, we can start by graphing the system of constraints and then finding the point that maximizes the objective function.
First, let's graph the constraints. The first constraint, x ≥ 0 and y ≥ 0, represents that we can't have negative quantities of printers. We can represent this as the area in the first quadrant of the graph.
The second constraint, y ≤ (1/5)x + 2, is a linear inequality. We can graph this inequality by graphing the line y = (1/5)x + 2 and shading the region below the line.
The third constraint, 5 ≤ y + x, is another linear inequality. We can graph this by graphing the line y + x = 5 and shading the region above the line.
Now that we have all the constraints graphed, we need to find the point that maximizes the objective function C = 7x - 3y. We can do this by evaluating the objective function at the vertices of the feasible region (the region that satisfies all the constraints) and choosing the vertex that gives the maximum value of the objective function.
By evaluating the objective function at each vertex, we can find that the maximum value occurs at the vertex (0,2).
Therefore, the answer is b. (0,2)
2. This problem is asking for the minimum cost, not the maximum profit like the previous problem. We can solve this problem by following a similar approach as before.
Again, let's start by graphing the system of constraints. The constraints are the same as in the previous problem.
Next, we need to find the point that minimizes the cost. The cost is given by the equation Cost = 237x + 122y.
By evaluating the cost equation at each vertex of the feasible region, we can find that the minimum cost occurs at the vertex (40,80).
Therefore, the answer is b. 40 of type A: 80 of type B.
3. This problem asks for the production levels that yield the maximum profit. To solve it, we will use the same approach as before.
Let's start by graphing the constraints. The first constraint, y ≤ x + 200, represents that the production of y can exceed x by no more than 200 units. We can graph this constraint as a line with a slope of 1 and a y-intercept of 200.
The second constraint, x + 2y ≤ 1600, is a linear inequality. We can graph this inequality by graphing the line x + 2y = 1600 and shading the region below the line.
Now, we need to find the point that maximizes the profit. The profit is given by the equation Profit = 14x + 22y - 900.
By evaluating the profit equation at each vertex of the feasible region, we can find that the maximum profit occurs at the vertex (400,600).
Therefore, the answer is a. x = 400; y = 600.
I hope that explanation helps! Let me know if you have any further questions.