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? Possible answers:
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
To minimize the cost while meeting the given conditions, we can set up a linear programming problem. Let's define the variables:
Let x be the number of type A printers ordered.
Let y be the number of type B printers ordered.
The cost, C, can be calculated as follows:
C = (cost of type A printer * number of type A printers) + (cost of type B printer * number of type B printers)
C = (237 * x) + (122 * y)
The profit, P, can be calculated as follows:
P = (profit per type A printer * number of type A printers) + (profit per type B printer * number of type B printers)
P = (22 * x) + (19 * y)
Now, let's consider the given constraints:
1) We can order no more than 120 printers this month:
x + y ≤ 120
2) We need to make at least $2,400 profit on the printers:
P ≥ 2400
(22 * x) + (19 * y) ≥ 2400
3) We must order at least one of each type of printer:
x ≥ 1
y ≥ 1
To minimize the cost, we need to minimize the objective function C= (237 * x) + (122 * y) subject to the given constraints.
Using linear programming techniques, we find that the solution is:
69 of type A : 51 of type B
Therefore, you should order 69 type A printers and 51 type B printers to minimize your cost while meeting the given conditions.
Let's assume you order X printers of type A and Y printers of type B.
The total cost of ordering X printers of type A is 237X, and the total profit is 22X.
The total cost of ordering Y printers of type B is 122Y, and the total profit is 19Y.
According to the problem, you need to order no more than 120 printers:
X + Y ≤ 120
And you need to make at least $2,400 profit:
22X + 19Y ≥ 2,400
Since you must order at least one of each type of printer, we have the additional conditions:
X ≥ 1
Y ≥ 1
To minimize the cost, we need to find the pair (X, Y) that satisfies all the conditions and minimizes the total cost, which is given by:
Total cost = 237X + 122Y
Now let's solve these inequalities step by step.
1. X + Y ≤ 120
X ≥ 1
Y ≥ 1
To visualize the feasible region, let's graph these inequalities on the XY plane.
Drawing the line X + Y = 120, we find the region below or on the line.
Next, draw the line X = 1, which is a vertical line at X = 1. Shade the region to the right of this line.
Lastly, draw the line Y = 1, which is a horizontal line at Y = 1. Shade the region below this line.
Now, find the overlapping region that satisfies all three inequalities. It is the corner of the feasible region where all three shaded regions intersect, which is the point (1,1).
2. 22X + 19Y ≥ 2,400
Substitute the values of X and Y from the point (1,1) into the inequality:
22(1) + 19(1) ≥ 2,400
41 ≥ 2,400
Since this condition is not satisfied, the point (1,1) is not feasible.
We need to find another corner of the feasible region to satisfy the inequality. The three remaining points are (1, 119), (119, 1), and (60, 60).
3. Total cost = 237X + 122Y
a) For point (1, 119):
Total cost = 237(1) + 122(119)
= 237 + 14,518
= 14,755
b) For point (119, 1):
Total cost = 237(119) + 122(1)
= 28,203 + 122
= 28,325
c) For point (60, 60):
Total cost = 237(60) + 122(60)
= 14,220 + 7,320
= 21,540
The minimum total cost is 14,755, corresponding to the order of 1 type A printer and 119 type B printers.
Therefore, the correct answer is:
1 of type A : 119 of type B
To solve this problem, we need to set up a system of constraints and objective function. Let's define:
Let x be the number of type A printers to be ordered.
Let y be the number of type B printers to be ordered.
Since we must order at least one of each type of printer, we have the constraint that both x and y must be greater than or equal to 1:
x ≥ 1
y ≥ 1
We can also see from the problem that the total number of printers ordered cannot exceed 120:
x + y ≤ 120
And the minimum profit requirement is $2,400:
22x + 19y ≥ 2,400
Now, let's solve this system of constraints and objective function to find the minimum cost.
To visualize this, we can draw a graph. We'll plot the feasible region (the region satisfying the constraints) and find the minimum value of the objective function within that region.
To do that, we start by graphing the lines corresponding to the constraints:
x = 1
y = 1
x + y = 120
We'll shade the region of the graph that satisfies all the constraints.
Next, we'll draw a line representing the objective function:
22x + 19y = 2,400
We want to find the lowest point (minimum) on this line within the feasible region. We can test the corner points of the feasible region to find which one gives the minimum cost.
If we calculate the cost for each possible corner point:
(69, 51): Cost = (69 × $237) + (51 × $122)
(40, 80): Cost = (40 × $237) + (80 × $122)
(51, 69): Cost = (51 × $237) + (69 × $122)
(80, 40): Cost = (80 × $237) + (40 × $122)
We can see that the minimum cost occurs when ordering 69 of type A printers and 51 of type B printers:
69 of type A: 69 × $237 = $16,353
51 of type B: 51 × $122 = $6,222
Total cost = $16,353 + $6,222 = $22,575
Therefore, the answer is:
69 of type A: 51 of type B