If x and y are two-digit integers such that x > 40 and y , which of the following is...

1. Translate the problem requirements: We need to find two integers where \(\mathrm{x}\) is a two-digit number greater than 40, \(\mathrm{y}\) is a two-digit number less than 70, and we want the largest possible value of their product \(\mathrm{xy}\).
2. Identify the constraints for maximum values: Determine the largest possible values for \(\mathrm{x}\) and \(\mathrm{y}\) given that \(\mathrm{x} > 40\), \(\mathrm{y} < 70\), and both must be two-digit integers.
3. Calculate the maximum product: Multiply the maximum possible values of \(\mathrm{x}\) and \(\mathrm{y}\) to get the largest possible product.
4. Compare with answer choices: Since the answer choices are spread apart by thousands, determine which choice is closest to our calculated maximum.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us in everyday language. We have two numbers, x and y, and we need to understand the rules for each:

For x: It must be a two-digit number (so between 10 and 99) AND it must be greater than 40. This means x can be any whole number from 41 up to 99.

For y: It must be a two-digit number (so between 10 and 99) AND it must be less than 70. This means y can be any whole number from 10 up to 69.

Our goal: We want to find the largest possible value when we multiply x and y together.

Process Skill: TRANSLATE - Converting the constraint language into clear mathematical boundaries

2. Identify the constraints for maximum values

To get the biggest possible product when multiplying two numbers, we simply want to use the biggest possible values for each number.

What's the largest possible x? Since x must be greater than 40 and must be a two-digit number, the largest value x can be is 99.

What's the largest possible y? Since y must be less than 70 and must be a two-digit number, the largest value y can be is 69.

Think of it this way: if you're trying to get the biggest area for a rectangle, you'd want to make both the length and width as large as possible within your constraints.

Process Skill: APPLY CONSTRAINTS - Systematically finding the maximum values within the given boundaries

3. Calculate the maximum product

Now we simply multiply our maximum values together:

Maximum product = \(99 \times 69\)

Let's calculate this step by step to keep it manageable:
\(99 \times 69 = 99 \times (70 - 1)\)
\(= 99 \times 70 - 99 \times 1\)
\(= 6,930 - 99\)
\(= 6,831\)

So our maximum possible value for xy is 6,831.

4. Compare with answer choices

Looking at our answer choices:

1. 700 - much too small
2. 2,800 - still too small
3. 4,000 - getting closer but still too small
4. 7,000 - very close to our calculated value of 6,831
5. 28,000 - much too large

Our calculated maximum of 6,831 is closest to choice (D) 7,000.

The difference is only 169, while the next closest choice (C) 4,000 is off by 2,831.

Final Answer

The answer is (D) 7,000. By maximizing both x and y within their constraints (\(\mathrm{x} = 99\), \(\mathrm{y} = 69\)), we get a product of 6,831, which is closest to 7,000 among the given choices.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint boundaries
Students often confuse whether the constraints are inclusive or exclusive. Since \(\mathrm{x} > 40\), some students might think x can equal 40, when it actually must be at least 41. Similarly, since \(\mathrm{y} < 70\), students might think y can equal 70, when it actually must be at most 69. This boundary confusion leads to using incorrect maximum values (\(\mathrm{x} = 100\) or \(\mathrm{y} = 70\)) right from the start.

2. Overlooking the two-digit requirement
Students may focus so heavily on the inequality constraints (\(\mathrm{x} > 40\) and \(\mathrm{y} < 70\)) that they forget both numbers must be two-digit integers. This could lead them to consider \(\mathrm{x} = 100\) as a valid option, since \(100 > 40\), without realizing that 100 is a three-digit number and therefore not allowed.

3. Not recognizing this as a straightforward maximization problem
Some students might overthink the problem and assume there's a trick or special case to consider. They might wonder if smaller values could somehow produce a larger product, or if there are other mathematical relationships to explore, when the solution is simply to maximize both variables within their constraints.

Errors while executing the approach

1. Arithmetic errors in multiplication
When calculating \(99 \times 69\), students might make computational mistakes. Common errors include incorrectly applying the distributive method (\(99 \times 70 - 99 \times 1\)) or making basic multiplication errors that lead to results like 6,831 becoming 6,381 or other variations.

2. Using incorrect maximum values due to constraint misunderstanding
If students misunderstood the constraints in phase 1, they'll execute the calculation with wrong values. For example, using \(\mathrm{x} = 100\) and \(\mathrm{y} = 70\) would give 7,000, or using \(\mathrm{x} = 40\) and \(\mathrm{y} = 70\) would give 2,800. These errors carry forward from the approach phase but manifest as execution errors.

Errors while selecting the answer

1. Selecting an exact match when approximation is required
Students who somehow arrive at exactly 7,000 (perhaps through calculation errors) might feel confident selecting choice (D) without realizing their calculation was wrong. The question asks for the "closest" value, indicating that approximation is expected, but students might not recognize this if their incorrect calculation yields a perfect match.

2. Poor judgment in proximity comparison
When comparing 6,831 to the answer choices, students might not clearly calculate the differences. They might see that 6,831 is between 4,000 and 7,000 and incorrectly choose 4,000 because "6 is closer to 4 than to 7" without properly calculating that \(|6,831 - 7,000| = 169\) while \(|6,831 - 4,000| = 2,831\).