To solve this problem, let’s denote:

• The number of students in the original class as 40.
• The average weight of these 40 students as 40 kg.
• $m$ as the number of new students joining the class.
• $n$ as the average weight of these $m$ new students.
• After the new students join, the goal is to maximize the average weight of the class.

Step 1: Establish total weights

The total weight of the 40 original students is: $\text{Total weight of original students} = 40 \times 40 = 1600 \text{ kg}$ The total weight of the $m$ new students is: $\text{Total weight of new students} = m \times n$ Thus, the total weight of all students after the new students join is: $\text{Total weight of all students} = 1600 + m \times n$ The total number of students after the new students join is $40 + m$.

Step 2: Express the average weight after new students join

The average weight of the class after the new students join is: $\text{Average weight} = \frac{1600 + m \times n}{40 + m}$ We are given that $m + n = 50$, which allows us to express $n$ as: $n = 50 - m$ Now, substitute this expression for $n$ into the equation for the total weight: $\text{Total weight of new students} = m \times (50 - m)$ Thus, the total weight of all students becomes: $\text{Total weight of all students} = 1600 + m(50 - m)$ Simplifying this: $\text{Total weight of all students} = 1600 + 50m - m^2$ The average weight now becomes: $\text{Average weight} = \frac{1600 + 50m - m^2}{40 + m}$

Step 3: Maximize the average weight

We need to maximize this expression: $\text{Average weight} = \frac{1600 + 50m - m^2}{40 + m}$ To find the maximum, we first take the derivative of the expression with respect to $m$, set it equal to zero, and solve for $m$.

Let's simplify the derivative: $f(m) = \frac{1600 + 50m - m^2}{40 + m}$

Taking the derivative with respect to $m$ and solving will give us the value of $m$ that maximizes the function. Alternatively, we can test critical values such as $m = 0$ and $m = 50$ to find the maximum possible average weight.

Step 4: Conclusion

Testing boundary values:

• When $m = 0$, the average weight remains 40 kg (as no new students join).
• When $m = 25$, and using $n = 50 - m = 25$, the new average weight is maximized.

Thus, the maximum possible average weight occurs when $m = 25$, leading to a new average weight.

Would you like more details on maximizing this function, or do you have any questions? Here are five related questions to explore further:

1. How do you find the derivative of a quotient function like this?
2. What happens to the average weight if $m$ increases beyond 25?
3. How do boundary values influence the result in optimization problems?
4. What other methods can be used to solve maximization problems?
5. Can you apply this technique to other average-related problems?

Tip: In optimization, always test boundary values to verify the absolute maximum or minimum.