Your furniture store sells two types of dining room tables. The first, type A, costs $265 and you make a $25 profit on each one. The second, type B, costs $100 and you make a $13 profit on each one. You can order no more than 40 tables this month, and you need to make at least $760 profit on them. If you must order at least one of each type of table, how many of each type of table should you order if you want to minimize your cost?
(1 point)
Responses
20 of type A; 20 of type B
20 of type A; 20 of type B
2 of type A; 38 of type B
2 of type A; 38 of type B
30 of type A; and 10 of type B
30 of type A; and 10 of type B
38 of type A; 2 of type B
To minimize the cost, we need to find the combination that meets the profit requirement while minimizing the cost. Let's assume we order x tables of type A and y tables of type B.
The profit from type A tables is $25 per table, so the total profit from type A tables is 25x.
The profit from type B tables is $13 per table, so the total profit from type B tables is 13y.
According to the problem, the total profit should be at least $760, so we have the inequality: 25x + 13y ≥ 760.
We are also limited to ordering no more than 40 tables, so the total number of tables should be less than or equal to 40: x + y ≤ 40.
To find the combination that minimizes the cost, we can consider the cost of each type of table. The cost of type A tables is $265 per table, so the total cost of type A tables is 265x. The cost of type B tables is $100 per table, so the total cost of type B tables is 100y.
Now, let's solve the problem using linear programming:
Objective function:
Minimize the cost: Cost = 265x + 100y
Constraints:
25x + 13y ≥ 760
x + y ≤ 40
x ≥ 1 (we must order at least one of each type)
Based on the given options, let's check each possibility:
1. 20 of type A; 20 of type B:
Profit: 20 * 25 + 20 * 13 = $900
Cost: 20 * 265 + 20 * 100 = $7500
Not meeting the profit requirement.
2. 2 of type A; 38 of type B:
Profit: 2 * 25 + 38 * 13 = $614
Cost: 2 * 265 + 38 * 100 = $4730
Not meeting the profit requirement.
3. 30 of type A; 10 of type B:
Profit: 30 * 25 + 10 * 13 = $835
Cost: 30 * 265 + 10 * 100 = $9050
Not meeting the profit requirement.
4. 38 of type A; 2 of type B:
Profit: 38 * 25 + 2 * 13 = $1024
Cost: 38 * 265 + 2 * 100 = $10570
Profits meet the requirement.
Therefore, the optimal order would be 38 of type A and 2 of type B.
To minimize the cost, let's assume you order x tables of type A and y tables of type B.
The cost of x tables of type A is 265x, and the profit is 25x.
The cost of y tables of type B is 100y, and the profit is 13y.
We have the following constraints:
1) x + y ≤ 40 (cannot order more than 40 tables)
2) 25x + 13y ≥ 760 (need to make at least $760 profit)
To find the optimal solution, we will use linear programming techniques.
The possible combinations of x and y satisfying the constraints are:
1) 20 of type A; 20 of type B
2) 2 of type A; 38 of type B
3) 30 of type A; and 10 of type B
4) 38 of type A; 2 of type B
To determine which combination minimizes the cost, we need more information. Could you provide additional criteria or preferences for selecting the optimal solution?
To solve this problem, we need to find the number of each type of table that will result in the minimum cost while also meeting the profit requirement.
Let's assume the number of type A tables ordered is "x" and the number of type B tables ordered is "y".
Given the cost and profit information, we can calculate the cost and profit for each type of table:
Cost of type A tables = $265
Profit of type A tables = $25
Cost of type B tables = $100
Profit of type B tables = $13
Now, we need to set up the constraints and objective function:
Constraints:
1. The total number of tables ordered cannot exceed 40:
x + y ≤ 40
2. The profit from the tables should be at least $760:
25x + 13y ≥ 760
Objective function:
We want to minimize the cost, which is given by:
Cost = (Cost of type A tables × Number of type A tables) + (Cost of type B tables × Number of type B tables)
Now, let's solve the problem using the method of linear programming:
1. Start by graphing the feasible region determined by the constraints x + y ≤ 40 and 25x + 13y ≥ 760.
2. Identify the corner points of the feasible region, which are the intersections of the constraint lines.
3. Calculate the cost at each corner point using the objective function.
4. Determine the corner point(s) that result in the minimum cost.
From the calculations, we find that the minimum cost occurs when we order 30 type A tables and 10 type B tables.
Therefore, the answer is:
30 of type A; and 10 of type B.