A farm supply store carries 50-lb bags of both grain pellets and grain mash for pig feed. It can store 600 bags of pig feed. At least twice as many of its customers prefer the mash to the pellets. The store buys the pellets for $3.75 per bag and sells them for $6.00. It buys the mash for $2.50 per bag and sells it for $4.00. If the store orders no more that $1400 worth of pig feed, how many bags of mash should the store order to make the most profit?
Answer: 400 bags
To find the optimal number of bags of mash that the store should order, we need to determine the number of bags of pellets and mash that maximize the store's profit. Let's break down the problem step by step:
1. Let's represent the number of bags of pellets as "x" and the number of bags of mash as "y."
2. From the given information, we know the following:
- The store can store a maximum of 600 bags of pig feed, so x + y ≤ 600.
- The number of customers preferring mash is at least twice the number preferring pellets, so y ≥ 2x.
- The cost of buying pellets is $3.75 per bag, and the selling price is $6.00 per bag.
- The cost of buying mash is $2.50 per bag, and the selling price is $4.00 per bag.
- The store wants to order pig feed worth no more than $1400, so the cost of buying pellets and mash should be less than or equal to $1400, which gives us the inequality 3.75x + 2.50y ≤ 1400.
3. Based on the above conditions, we can set up the following optimization problem:
Maximize profit = (6x - 3.75x) + (4y - 2.50y) = 2.25x + 1.50y, subject to the constraints:
- x + y ≤ 600,
- y ≥ 2x,
- 3.75x + 2.50y ≤ 1400.
4. To solve this optimization problem, we need to find the maximum value of the objective function 2.25x + 1.50y within the feasible region determined by the constraints.
5. The solution to this problem can be found using graphical methods or linear programming. However, for simplicity, we can solve it by considering the given conditions.
- Since y ≥ 2x, let's try an initial value of x = 100. This gives us y = 200.
- Substituting the values in the objective function, we get profit = 2.25(100) + 1.50(200) = 225 + 300 = $525.
- Now, let's increase the value of x by 100 and calculate the corresponding value of y.
- For x = 200, y = 400 (y ≥ 2x is maintained).
- Substituting the values, we get profit = 2.25(200) + 1.50(400) = 450 + 600 = $1050.
- Continuing this process, when x = 300, we get y = 600 (y ≥ 2x).
- Profit = 2.25(300) + 1.50(600) = 675 + 900 = $1575.
- However, note that the constraint x + y ≤ 600 is violated since x + y = 300 + 600 = 900 > 600.
- To satisfy all the constraints, we can try reducing x by 100, giving us x = 200.
- Then, y = 400 (y ≥ 2x).
- Substituting the values, we get profit = 2.25(200) + 1.50(400) = 450 + 600 = $1050.
- As we further reduce x to 100 (keeping y ≥ 2x), the profit decreases to $525.
- Therefore, the maximum profit of $1050 is achieved when x = 200 and y = 400.
Therefore, the store should order 400 bags of mash to make the most profit.