Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A manufacturer of car radios ships them to retailers in cartons of $n$ radios. The profit per radio is $\$59.50$, minus shipping cost of $\$25$ per carton, so the profit is $59.5n-25$ dollars per carton. To promote sales by assuring high quality, the manufacturer promises to pay the retailer $\$200X^2$ if $X$ radios in the carton are defective. Suppose radios are produced independently and that $5\%$ of radios are defective. How many radios should be packed per carton to maximize expected net profit per carton?

My solution:

Suppose that $n$ radios are packed into a carton. Then, the expected profit is clearly $$\mu = 59.5n-25-200(0.05n)^2$$ Simplifying we get $$\mu = 59.5n-25-0.5n^2$$

We want to find a maximum, so we find a derivative:


Clearly there is just one maximum, so set $\mu'=0$ we find that $$n^2=59.5$$ and thus $$n\approx 7.7136\dots$$This however seems to be wrong. The textbook gives an answer of exactly $50$ in the solution without any explanation of the steps.

share|cite|improve this question
up vote 2 down vote accepted

$X$ is binomially distributed with $n$ trials and success probability $0.05$. For any random variable $X$ whose mean and variance exist, \begin{eqnarray*} {\Bbb E}X^2&=&({\Bbb E}X)^2+\text{Var } X, \end{eqnarray*} and since the mean and variance of a binomial distribution with $n$ trials and success probability $p$ are $np$ and $np(1-p)$, in this case this equals \begin{eqnarray*} &&(0.05 n)^2+n\cdot 0.05 \cdot (1-0.05)\\ &=& 0.0025 n^2+0.0475 n \end{eqnarray*} and the expected profit is \begin{eqnarray*} &\ &59.5n-25-200{\Bbb E}X^2\\ &=&59.5n-25-0.5n^2-9.5n\\ &=&1225-\frac 12 (n-50)^2, \end{eqnarray*} which is maximized at $n=50$.

share|cite|improve this answer

The expected loss isn't just $(0.05n)^2$, you must compute its average the hard way ($\mathbb{E}(x^2) \ne (\mathbb{E}(x))^2$ in general!). If the probability of $n$ ones broken is $p_n$, your expected loss due to breakage is $$ \sum_{n \ge 0} p_n \cdot 200 n^2 = 200 \sum_{n \ge 0} p_n n^2 $$ Presumably you can start assuming that the number of radios in each box isn't limited, that should give a starting point. Once you have an approximate number of radios per box, refine.

share|cite|improve this answer

The profit $Y$ is given by $$Y=59.5n -25-200 X^2,$$ where $X$ is the number of defectives.

The number of defectives has binomial distribution, with $p=0.05$.

We want to find $E(X^2)$. There are various ways to do this. One way is to recall that the relevant binomial has mean $np$ and variance $np(1-p)$. But the variance of $X$ is $E(X^2)-(E(X))^2. So E(X^2)=np(1-p) +n^2p^2$.

There are various other ways to find $E(X^2)$.

So we are maximizing $59.5n-25-200np(1-p)-200n^2p^2$, which in this case is $-(0.5n^2 -50n +25)$.

For the maximization, complete the square, or use calculus.

The answer does turn out to be exactly $50$.

Remark: You asked why your answer was wrong. One way was uninteresting. You differentiated $59.5n-0.5n^2$ and got $59.5-n^2$ instead of the correct $59.5-n$.

That would have given an answer of $59.5$, say $60$-ish.

The actual answer is less, and for an interesting reason. The penalty for bad radios is $200X^2$. This is quite sensitive to large values of $X$. Roughly speaking that explains why the optimal number of radios is smaller than the $59.5$ given by your method.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.