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A company wants to buy computers and printers for a new branch office, and the number of computers can be at most \(3\) times the number of printers. Computers cost \(\$1{,}500\) each, and printers cost \(\$300\) each. What is the greatest number of computers that the company can buy if it has a total of \(\$9{,}100\) to spend on computers and printers?
Let's start by understanding what we're dealing with in everyday terms. The company has \(\text{\$9,100}\) to spend on computers and printers for a new office. Think of this like having a shopping budget - we want to buy as many computers as possible, but we have rules to follow.
The first rule is about cost: computers are expensive at \(\text{\$1,500}\) each, while printers are much cheaper at \(\text{\$300}\) each. The second rule is about the ratio: we can't go crazy buying computers - we can have "at most 3 times" the number of printers. This means if we buy 1 printer, we can have up to 3 computers; if we buy 2 printers, we can have up to 6 computers, and so on.
In mathematical terms:
• Let \(\mathrm{C}\) = number of computers, \(\mathrm{P}\) = number of printers
• Budget constraint: \(1500\mathrm{C} + 300\mathrm{P} \leq 9100\)
• Ratio constraint: \(\mathrm{C} \leq 3\mathrm{P}\)
• Goal: Maximize \(\mathrm{C}\)
Process Skill: TRANSLATE - Converting the business language "at most 3 times" into the mathematical constraint \(\mathrm{C} \leq 3\mathrm{P}\)
Now let's think about which rule will actually stop us from buying more computers. Will it be running out of money, or will it be the ratio rule?
Let's imagine we could ignore the ratio rule for a moment. With \(\text{\$9,100}\) and computers costing \(\text{\$1,500}\) each, we could theoretically buy \(9100 \div 1500 = \text{about } 6\) computers if we bought no printers. But wait - if we buy no printers, then the ratio rule says we can have at most \(3 \times 0 = 0\) computers! So we definitely need some printers.
Let's try the opposite extreme. If we wanted to satisfy the ratio rule easily, we could buy lots of printers. But printers cost money too, and every printer we buy means less money for computers.
This tells us we need to find the sweet spot where both constraints work together.
Process Skill: INFER - Recognizing that both constraints will interact and we need to find the optimal balance
Since we want to maximize computers, let's work backwards from the answer choices and see what works. The highest option is 6 computers, so let's start there.
Testing 6 computers:
If \(\mathrm{C} = 6\), then from the ratio rule: \(6 \leq 3\mathrm{P}\), so \(\mathrm{P} \geq 2\). Let's try \(\mathrm{P} = 2\).
Cost check: \(6 \times \text{\$1,500} + 2 \times \text{\$300} = \text{\$9,000} + \text{\$600} = \text{\$9,600}\)
But we only have \(\text{\$9,100}\)! This is too expensive.
Testing 5 computers:
If \(\mathrm{C} = 5\), then from the ratio rule: \(5 \leq 3\mathrm{P}\), so \(\mathrm{P} \geq 5/3 = 1.67\). Since we need whole printers, \(\mathrm{P} \geq 2\).
Cost check: \(5 \times \text{\$1,500} + 2 \times \text{\$300} = \text{\$7,500} + \text{\$600} = \text{\$8,100}\)
This costs \(\text{\$8,100}\), which is less than \(\text{\$9,100}\). Great! This works.
Let's double-check: Do we have \(5 \leq 3 \times 2\)? Yes, \(5 \leq 6\) ✓
Do we have enough money? Yes, \(\text{\$8,100} \leq \text{\$9,100}\) ✓
Can we do better than 5 computers?
Maybe we could buy more printers to allow more computers? Let's try 5 computers with 3 printers:
Cost: \(5 \times \text{\$1,500} + 3 \times \text{\$300} = \text{\$7,500} + \text{\$900} = \text{\$8,400}\)
This still fits our budget, but we're not increasing computers.
Let's try 6 computers with 2 printers again, but check if we could use 3 printers:
6 computers with 3 printers: \(6 \times \text{\$1,500} + 3 \times \text{\$300} = \text{\$9,000} + \text{\$900} = \text{\$9,900}\)
Still over budget.
Process Skill: APPLY CONSTRAINTS - Systematically checking both the ratio and budget constraints for each scenario
Our best solution so far is 5 computers and 2 printers. Let's make sure this is truly optimal by checking if we missed anything.
Final verification for 5 computers, 2 printers:
• Ratio constraint: \(5 \leq 3 \times 2 = 6\) ✓
• Budget constraint: \(5 \times \text{\$1,500} + 2 \times \text{\$300} = \text{\$8,100} \leq \text{\$9,100}\) ✓
• Remaining budget: \(\text{\$9,100} - \text{\$8,100} = \text{\$1,000}\)
Can we use that extra \(\text{\$1,000}\)? We could buy more printers, but that doesn't help us get more computers. We've already established that 6 computers requires too much money.
Therefore, the maximum number of computers we can buy is 5.
The greatest number of computers that the company can buy is 5.
This corresponds to answer choice (D) 5.
Our optimal solution uses 5 computers and 2 printers, costing \(\text{\$8,100}\) total, which satisfies both the budget constraint and the ratio constraint that computers cannot exceed 3 times the number of printers.
Students often confuse "at most 3 times the number of printers" with "exactly 3 times" or "at least 3 times." This leads them to write \(\mathrm{C} = 3\mathrm{P}\) instead of \(\mathrm{C} \leq 3\mathrm{P}\), which fundamentally changes the problem. The word "at most" means "no more than," so if there are 2 printers, you can have 1, 2, 3, 4, 5, or 6 computers, but not 7 or more.
Some students might try to minimize cost or maximize the total number of items (computers + printers) instead of focusing on maximizing computers specifically. The question clearly asks for "the greatest number of computers," but students sometimes get distracted by wanting to use the entire budget efficiently rather than achieving the actual goal.
When setting up the mathematical relationships, students might forget that you can't buy fractional computers or printers. This becomes critical when working with the constraint \(\mathrm{C} \leq 3\mathrm{P}\), especially when solving for minimum printer requirements like \(\mathrm{P} \geq 5/3\), where students must round up to \(\mathrm{P} = 2\).
With relatively large numbers (\(\text{\$1,500}\) and \(\text{\$300}\)), students frequently make arithmetic mistakes when calculating total costs. For example, computing \(5 \times \text{\$1,500} + 2 \times \text{\$300}\) might result in errors like \(\text{\$7,500} + \text{\$600} = \text{\$8,000}\) instead of \(\text{\$8,100}\), leading to incorrect budget assessments.
Students might test scenarios without systematically checking both constraints. For instance, when testing 6 computers, they might only check if \(6 \leq 3\mathrm{P}\) with \(\mathrm{P} = 2\) (which gives \(6 \leq 6\), seeming valid) but forget to verify the budget constraint, missing that this actually costs \(\text{\$9,600}\), which exceeds the \(\text{\$9,100}\) budget.
When working with \(\mathrm{C} \leq 3\mathrm{P}\), students often make errors in rearranging to find \(\mathrm{P} \geq \mathrm{C}/3\). For 5 computers, they need \(\mathrm{P} \geq 5/3 = 1.67\), but students might incorrectly conclude that \(\mathrm{P} = 1\) is sufficient, or they might round 1.67 down instead of up to the nearest whole number.
Students might notice that their solution (5 computers, 2 printers costing \(\text{\$8,100}\)) leaves \(\text{\$1,000}\) unused and feel this can't be optimal. They might then try to force a solution that uses more of the budget, even if it violates constraints, and select a higher answer choice thinking "more expensive must be better."
During the solution process, students calculate various numbers: costs (\(\text{\$8,100}\), \(\text{\$9,600}\)), constraint values (\(3\mathrm{P} = 6\)), and item quantities. Some students might accidentally select the number of printers (2) or mistake other calculated values for the final answer instead of clearly identifying that the question asks for the number of computers (5).