Question

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situation 2 : la correction de monsieur ben
monsieur ben, enseignant en mathématiques, doit corriger les examens de ses élèves. il corrige le matin en prenant son café et en soirée en prenant son bain. en un mois, monsieur ben corrige dans son coin de 3 à 9 fois et il corrige au maximum 14 fois en prenant son café. il a tendance à corriger au moins 2 fois de plus en prenant son café quen prenant son bain. finalement, chaque mois, il corrige au moins 12 fois.
sachant quil corrige en moyenne 7 copies dexamen chaque fois quil prend son bain et 4 copies chaque fois quil prend un café, quel est le nombre maximum de copies quil peut espérer corriger en un mois ?

Explanation:

To solve this problem, we need to determine the maximum number of exam copies Mr. Ben can correct in a month by considering the constraints on the number of times he corrects while having coffee (\(x\)) and while having a bath (\(y\)).

Step 1: Define the Variables and Constraints

Let \(x\) be the number of times he corrects while having coffee, and \(y\) be the number of times he corrects while having a bath.

The constraints are:

1. \(3 \leq y \leq 9\) (he corrects between 3 and 9 times while having a bath)
2. \(x \leq 14\) (he corrects at most 14 times while having coffee)
3. \(x \geq y + 2\) (he tends to correct at least 2 more times while having coffee than while having a bath)
4. \(x + y \geq 12\) (he corrects at least 12 times in total each month)

The objective function to maximize is the total number of copies corrected: \(T = 4x + 7y\) (since he corrects 4 copies per coffee session and 7 copies per bath session).

Step 2: Analyze the Constraints to Find Feasible Values

From the constraint \(x \geq y + 2\) and \(3 \leq y \leq 9\), we can express \(x\) in terms of \(y\). Also, we know \(x \leq 14\) and \(x + y \geq 12\).

Let's consider the upper bound for \(y\) (since \(y\) has a higher coefficient in the objective function, increasing \(y\) will have a greater impact on the total number of copies). The maximum value of \(y\) is 9.

For \(y = 9\):

• From \(x \geq y + 2\), we get \(x \geq 9 + 2 = 11\)
• From \(x \leq 14\), \(x\) can be from 11 to 14
• From \(x + y \geq 12\), for \(y = 9\), \(x + 9 \geq 12 \implies x \geq 3\), which is already satisfied by \(x \geq 11\)

Step 3: Calculate the Total for Feasible \(x\) and \(y\)

We need to maximize \(T = 4x + 7y\). Since \(y = 9\) (maximum \(y\)) and \(x\) is as large as possible (since \(x\) has a positive coefficient), we take the maximum \(x\) possible for \(y = 9\), which is \(x = 14\) (since \(x \leq 14\) and \(x \geq 11\) for \(y = 9\)).

Step 4: Verify the Constraints and Calculate the Total

Check the constraints for \(x = 14\) and \(y = 9\):

• \(3 \leq 9 \leq 9\) (satisfied)
• \(14 \leq 14\) (satisfied)
• \(14 \geq 9 + 2 = 11\) (satisfied)
• \(14 + 9 = 23 \geq 12\) (satisfied)

Now calculate the total number of copies:
\[
T = 4x + 7y = 4(14) + 7(9) = 56 + 63 = 119
\]

Step 5: Check if a Higher Total is Possible

Let's check if a higher \(y\) is possible, but \(y\) is already at its maximum of 9. Let's check if a lower \(y\) with a higher \(x\) could give a higher total. For example, if \(y = 8\), then \(x \geq 10\), and the maximum \(x = 14\):
\[
T = 4(14) + 7(8) = 56 + 56 = 112
\]
Which is less than 119. Similarly, for \(y = 7\), \(x = 14\):
\[
T = 4(14) + 7(7) = 56 + 49 = 105
\]
Which is also less than 119.

Thus, the maximum total occurs when \(x = 14\) and \(y = 9\).

\[
\boxed{119}
\]

Answer:

To solve this problem, we need to determine the maximum number of exam copies Mr. Ben can correct in a month by considering the constraints on the number of times he corrects while having coffee (\(x\)) and while having a bath (\(y\)).

Step 1: Define the Variables and Constraints

Let \(x\) be the number of times he corrects while having coffee, and \(y\) be the number of times he corrects while having a bath.

The constraints are:

1. \(3 \leq y \leq 9\) (he corrects between 3 and 9 times while having a bath)
2. \(x \leq 14\) (he corrects at most 14 times while having coffee)
3. \(x \geq y + 2\) (he tends to correct at least 2 more times while having coffee than while having a bath)
4. \(x + y \geq 12\) (he corrects at least 12 times in total each month)

The objective function to maximize is the total number of copies corrected: \(T = 4x + 7y\) (since he corrects 4 copies per coffee session and 7 copies per bath session).

Step 2: Analyze the Constraints to Find Feasible Values

From the constraint \(x \geq y + 2\) and \(3 \leq y \leq 9\), we can express \(x\) in terms of \(y\). Also, we know \(x \leq 14\) and \(x + y \geq 12\).

Let's consider the upper bound for \(y\) (since \(y\) has a higher coefficient in the objective function, increasing \(y\) will have a greater impact on the total number of copies). The maximum value of \(y\) is 9.

For \(y = 9\):

• From \(x \geq y + 2\), we get \(x \geq 9 + 2 = 11\)
• From \(x \leq 14\), \(x\) can be from 11 to 14
• From \(x + y \geq 12\), for \(y = 9\), \(x + 9 \geq 12 \implies x \geq 3\), which is already satisfied by \(x \geq 11\)

Step 3: Calculate the Total for Feasible \(x\) and \(y\)

We need to maximize \(T = 4x + 7y\). Since \(y = 9\) (maximum \(y\)) and \(x\) is as large as possible (since \(x\) has a positive coefficient), we take the maximum \(x\) possible for \(y = 9\), which is \(x = 14\) (since \(x \leq 14\) and \(x \geq 11\) for \(y = 9\)).

Step 4: Verify the Constraints and Calculate the Total

Check the constraints for \(x = 14\) and \(y = 9\):

• \(3 \leq 9 \leq 9\) (satisfied)
• \(14 \leq 14\) (satisfied)
• \(14 \geq 9 + 2 = 11\) (satisfied)
• \(14 + 9 = 23 \geq 12\) (satisfied)

Now calculate the total number of copies:
\[
T = 4x + 7y = 4(14) + 7(9) = 56 + 63 = 119
\]

Step 5: Check if a Higher Total is Possible

Let's check if a higher \(y\) is possible, but \(y\) is already at its maximum of 9. Let's check if a lower \(y\) with a higher \(x\) could give a higher total. For example, if \(y = 8\), then \(x \geq 10\), and the maximum \(x = 14\):
\[
T = 4(14) + 7(8) = 56 + 56 = 112
\]
Which is less than 119. Similarly, for \(y = 7\), \(x = 14\):
\[
T = 4(14) + 7(7) = 56 + 49 = 105
\]
Which is also less than 119.

Thus, the maximum total occurs when \(x = 14\) and \(y = 9\).

\[
\boxed{119}
\]