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
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To find out how many bags of mash the store should order to make the most profit, we need to do a little math.
Let's assume the store orders x bags of mash. Since at least twice as many customers prefer the mash to the pellets, we know that the number of bags of pellets ordered will be at most half of x.
The cost of buying x bags of mash will be 2.5x dollars, while the cost of buying at most (1/2)x bags of pellets will be at most (1/2)x * 3.75 dollars.
Adding these costs together, we get: 2.5x + (1/2)x * 3.75.
Now, since the store can order no more than $1400 worth of pig feed, we can set up an equation:
2.5x + (1/2)x * 3.75 ≤ 1400
Simplifying this equation, we have:
2.5x + 1.875x ≤ 1400
Combining like terms, we get:
4.375x ≤ 1400
Dividing by 4.375, we find that x ≤ 320.
So, the store can order at most 320 bags of mash. However, we also know that the store can store a total of 600 bags of pig feed.
Since the store needs to make the most profit, it should order as many bags of mash as possible while still being able to store at most 600 bags in total. So, the store should order x = 320 bags of mash (the maximum amount) and (1/2)x = 160 bags of pellets.
Therefore, the store should order 320 bags of mash to make the most profit.
To find out how many bags of mash the store should order to make the most profit, let's work through the problem step-by-step.
Let's assume the number of bags of pellets the store purchases is p, and the number of bags of mash is m.
According to the given information:
1. The store can store a maximum of 600 bags of pig feed, so the total number of bags of pig feed should be p + m ≤ 600.
2. At least twice as many customers prefer mash to pellets, so m ≥ 2p.
3. The cost of each bag of pellets is $3.75, and the store buys p bags of pellets, so the cost of pellets is 3.75p (in dollars).
4. The cost of each bag of mash is $2.50, and the store buys m bags of mash, so the cost of mash is 2.50m (in dollars).
5. The store can order a maximum value of $1400 worth of pig feed, so the total cost of pellets and mash should be 3.75p + 2.50m ≤ 1400.
To maximize profit, the store needs to sell the most bags of feed at the highest possible price. The profit from selling pellets is $6.00 - $3.75 = $2.25 per bag, while the profit from selling mash is $4.00 - $2.50 = $1.50 per bag.
The objective is to find the number of bags of mash (m) that maximizes the profit.
Let's write the profit equation for selling both products:
Total Profit = Profit from pellets + Profit from mash
Total Profit = ($6.00 - $3.75)p + ($4.00 - $2.50)m
Total Profit = $2.25p + $1.50m
We need to maximize this objective function, subject to the constraints mentioned earlier:
p + m ≤ 600, and
3.75p + 2.50m ≤ 1400.
Now we can apply linear programming techniques to find the optimal solution. However, one way to derive a solution without complex mathematical calculations is to use logical reasoning.
If the store can store a maximum of 600 bags of pig feed, and the number of bags of mash (m) must be at least twice the number of bags of pellets (p), then the maximum possible value for m is 400 (twice the maximum possible value for p, which is 200). If we substitute these values into the objective function, keeping in mind that p + m should be less than or equal to 600, we get:
Total Profit = $2.25p + $1.50m
Total Profit = $2.25(200) + $1.50(400)
Total Profit = $450 + $600
Total Profit = $1050
Now, let's check whether these values satisfy the constraints:
p + m ≤ 600
200 + 400 ≤ 600
600 ≤ 600 (satisfied)
3.75p + 2.50m ≤ 1400
3.75(200) + 2.50(400) ≤ 1400
750 + 1000 ≤ 1400
1750 ≤ 1400 (not satisfied)
It seems that the total cost exceeds the maximum allowed value of $1400. Therefore, the previous solution is invalid.
To find an alternative solution, we can consider reducing the number of bags of mash to bring the total cost within the allowed limit. Let's suppose we buy 400 bags of mash:
Total Profit = $2.25p + $1.50m
Total Profit = $2.25(200) + $1.50(400)
Total Profit = $450 + $600
Total Profit = $1050
Let's check the constraints again:
p + m ≤ 600
200 + 400 ≤ 600
600 ≤ 600 (satisfied)
3.75p + 2.50m ≤ 1400
3.75(200) + 2.50(400) ≤ 1400
750 + 1000 ≤ 1400
1750 ≤ 1400 (not satisfied)
Even in this case, we find that the total cost exceeds the allowed maximum value.
Therefore, we need to reduce the number of bags of mash even further. Since the maximum value for p is 200 (given that the number of bags of mash should be at least twice the number of bags of pellets), let's try buying 200 bags of mash:
Total Profit = $2.25p + $1.50m
Total Profit = $2.25(200) + $1.50(200)
Total Profit = $450 + $300
Total Profit = $750
Let's check the constraints again:
p + m ≤ 600
200 + 200 ≤ 600
400 ≤ 600 (satisfied)
3.75p + 2.50m ≤ 1400
3.75(200) + 2.50(200) ≤ 1400
750 + 500 ≤ 1400
1250 ≤ 1400 (satisfied)
Now we see that the total cost is within the allowed limit.
Therefore, the maximum profit can be achieved by ordering 200 bags of pellets and 200 bags of mash.
To determine the number of bags of mash the store should order to maximize profit, we need to analyze the cost and revenue associated with each type of feed.
Let's start by calculating the profit from selling one bag of each feed:
For grain pellets:
Cost price = $3.75
Selling price = $6.00
Profit per bag of pellets = Selling price - Cost price = $6.00 - $3.75 = $2.25
For grain mash:
Cost price = $2.50
Selling price = $4.00
Profit per bag of mash = Selling price - Cost price = $4.00 - $2.50 = $1.50
Next, we need to consider the demand for each type of feed. The problem states that at least twice as many customers prefer the mash. Let's call the number of bags of mash ordered "m" and the number of bags of pellets ordered "p."
In this case, the number of bags of pellets ordered must be less than or equal to half the number of bags of mash ordered:
p ≤ m/2
In addition, the total number of bags ordered must not exceed the storage capacity of 600 bags:
p + m ≤ 600
Now let's consider the budget constraint. The store cannot order more than $1400 worth of pig feed.
The cost of the pellets equals the cost price multiplied by the number of bags:
3.75p ≤ 1400
The cost of the mash equals the cost price multiplied by the number of bags:
2.5m ≤ 1400
Now we can solve these inequalities to find the optimal solution.
Let's consider the constraints and calculate the maximum profit for different numbers of bags of mash.
Substituting p = m/2 into the cost constraint for pellets:
3.75(m/2) ≤ 1400
1.875m ≤ 1400
m ≤ 1400/1.875
m ≤ 746.67
Since we cannot have a fractional number of bags, we can round m down to its nearest whole number, making m = 746 for our purpose.
Now let's calculate the profit for m = 746:
Profit from pellets = 2.25 × (m/2)
Profit from mash = 1.5 × m
Next, let's calculate the total cost for m = 746 using the cost constraint for mash:
2.5 × 746 ≤ 1400 (this is already satisfied)
Finally, let's check whether this solution satisfies the storage constraint:
p + m ≤ 600
p + 746 ≤ 600
p ≤ 600 - 746
p ≤ -146
Since p cannot be negative, the solution of m = 746 is not valid.
Let's now consider m = 400, which is a whole number and ensures that there are enough pellets available:
p ≤ 400/2
p ≤ 200
We can confirm that the storage constraint is satisfied for m = 400:
p + 400 ≤ 600
p ≤ 600 - 400
p ≤ 200
Now let's calculate the profit for m = 400:
Profit from pellets = 2.25 × 200 = $450
Profit from mash = 1.5 × 400 = $600
The maximum profit is achieved when ordering 400 bags of mash, resulting in a profit of $600.