To minimize the cost, we need to maximize the profit. Let's denote the number of type A tables as A and the number of type B tables as B.
The profit from type A tables is given by: $25 x A
The profit from type B tables is given by: $13 x B
We are given that A + B = 40 (the total number of tables cannot exceed 40) and $25A + $13B ≥ $760 (we need to make at least $760 in profit).
To solve this problem, we can use a linear programming approach.
Since we need to order at least one of each type of table, we can assume that A and B are integers greater than or equal to 1.
We can set up the following linear programming problem:
Objective function: Minimize the cost, which in this case is the total cost of tables: 265A + 100B
Subject to:
A + B = 40 (the total number of tables cannot exceed 40)
25A + 13B ≥ 760 (we need to make at least $760 in profit)
A ≥ 1 (we need to order at least one of type A)
B ≥ 1 (we need to order at least one of type B)
A, B ≥ 0 (the number of tables cannot be negative)
Solving this linear programming problem, we find that the optimal solution is:
2 of type A tables and 38 of type B tables.
Therefore, the correct answer is:
2 of type A tables and 38 of type B tables.
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)Responses20 of type A; 20 of type B20 of type A; 20 of type B2 of type A; 38 of type B2 of type A; 38 of type B30 of type A; and 10 of type B30 of type A; and 10 of type B38 of type A; 2 of type B38 of type A; 2 of type B
1 answer
1 answer