probability problem in Binomial distribution The question:

A newsboy purchases papers at 12 cents and sells them at 16 cents.  However, he is not allowed to return unsold papers.  Suppose that his daily demand follows a Binomial distribution with n=10, p=1/3.  His mother suggests him to purchase 6 papers per day but his girlfriend suggests him to purchase 4 papers per day.  Who should the boy listen to so that he can obtain a higher expected profit?  

I am confused by this question.I tried the following:  
Define the probability distribution function by 
$$f(x)=\sum_{i=0}^{10}\binom{10}{x}p^x(1-p)^{10-x}$$ 
then define $u(x)=0.04x+(n-x)(-0.12)$.  
For $n=6$, find $\sum_{x=0}^6u(x)f(x)$;for $n=10$, find $\sum_{x=0}^{10}u(x)f(x)$,then compare the two sums.  But I do not quite understand the question. Could anyone help?
 A: If the boy purchases $n\leq10$ papers and the demand is $X$ then if $X>n$ he will gain $0.04n$ and if $X\leq n$ he will gain $0.04X-0.12(n-X)$.
That gives expectation: $$E_n:=\sum_{k=0}^n\left[0.04k-0.12(n-k)\right]P(X=k)+\sum_{k=n+1}^{10}0.04nP(X=k)$$
where $X$ has binomial distribution with parameters $10$ and $p=\frac13$
The question to be answered is actually: do we have $E_4>E_6$ or $E_6>E_4$?
I leave that to you.
A: If a newsboy purchases $k$ papers and sells $i$ papers ($i\leqslant k$), his profit is
$$16i - 12k$$ so the expected profit
is
$$\sum_{i=0}^k\binom{n}{i}p^i(1-p)^{n-i}\cdot (16i - 12k)+\sum_{i=k+1}^n\binom{n}{i}p^i(1-p)^{n-i}\cdot 4k$$
where $n=10$, $p=1/3$.
A: If you buy $n$ newspapers, and the demand that day is more than $n$, you will still make profit $u(n)$. Because you are currently only summing $u(x) f(x)$ up to $x=n$, and no further, you are ignoring the profits from these cases where demand exceeds supply. Instead of $\sum_{x=0}^n u(x) f(x)$, you should look at $\sum_{x=0}^{10} u(\min\{x,n\}) f(x)$.
Also, your expression for $f(x)$ has a summation over $i$ that isn't doing anything. It should be removed.
