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I am stuck up with this simple problem related to averages.Please help me out in explaining the complete solution to this problem.

Consider a class of 40 students whose average weight is $40$ kgs. $m$ new students join this class whose average weight is $n$ kgs. If it is known that $m + n = 50$, what is the maximum possible average weight of the class now?

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    *Hint*: The new total weight is $(1600+mn)\,\text{kg}$ – 2017-02-27
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    yeh i know that – 2017-02-27
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    i need the maximum possible average – 2017-02-27
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    guys plz help me out – 2017-02-27
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    Is it for a Calculus class? – 2017-02-27
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    yes kind of.. plz try to help me out friends – 2017-02-27
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    *More hint*: the new average weight is $\big(1600+m(50-m)\big)/(40+m)$ – 2017-02-27
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    could u plzz elaborate @dxiv – 2017-02-27

5 Answers 5

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My simple approach to the question:

Every new joinee should contribute to maximize average weight.

Therefore contribution if $1$ student joins $= 1 \times 9 = 9$ kg

And this will be maximum when both the variable is equal i.e., $5 \times 5 = 25$

Therefore maximum average weight $= 40 + \frac{25}{45} \approx 40.56$ kg

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Hint: Find the new total weight (which you claim you can find), divide by $40+m$, find the $m$ maximizing the expression!

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    plz show me out the maximizing part – 2017-02-27
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    i ain't getting the solution – 2017-02-27
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The new average is given by:

$$NA=\frac{40\cdot 40+m\cdot n}{40+m}$$

Once $m+n=50→n=50-m$ then

$$NA=f(m)=\frac{1600+50m-m^2}{40+m}=90-m-\frac{2000}{40+m}$$

In order to find the maximum you can do $f'(m)=0$.

Can you finish?

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After the join, the new sum is

\begin{align*} &(40)(40) + mn\\[6pt] =\; &1600 + mn\\[6pt] =\; &1600 + m(50-m) \end{align*}

and the new average is

$$f(m) = \frac{1600 + m(50-m)}{40 + m}$$

We want to maximize $f(m)$, for positive integer values of $m$. Taking the derivative, we get

$$f^{\prime}(m) = \frac{-m^2 - 80m + 400}{\left(m + 40\right)^2}$$

which has two real roots, but only one positive real root, $\,r = -40 + 20\sqrt{5} \approx 4.72$.

From the algebraic form of $f^{\prime}(m)$, it follows that

  • $0 < m < r \implies f^{\prime}(m) > 0$
  • $m > r \implies f^{\prime}(m) < 0$

Since $m$ is required to be a positive integer, and $4 < r < 5$, it follows that the optimal $m$ must be either $4$ or $5$.

Comparing $f(4)$ and $f(5)$, we find

$$f(4) = \frac{446}{11} < f(5) = \frac{365}{9}$$

It follows that the maximum possible new average is

$$f(5) = \frac{365}{9} = 40 + \frac{5}{9} \approx 40.56$$

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m+n = 50

The average weight with the new students will be

$AvW = \frac {40\cdot 40 + m\cdot n}{40+m}$

maximize $AvW$ constrained by $m+n = 50 $n = 50-m$

$AvW = \frac {40\cdot 40 + m\cdot (50-m)}{40+m}$

Will be optimized when $\frac {d}{dm} AvW = 0$