# Thread: Net profit Expectation Question

1. ## Net profit Expectation Question

Hi all,

I am having trouble with the problem:

A manufacturer ships car radios in cartons of n radios. The profit per radio is 59.50 per radio, and it costs 25 dollars to ship a carton. To promote sales and quality assurance, the retailed promises to pay the buyer 200X^2 for X radios that are defective. Suppose radios are produced independently and that 5% of radios are defective. How many radios should be placed per carton to maximize net profit.

So I have that the net profit described by (59.50*n) - 25 - 200 (0.05n)^2
where n is the number of boxes in a carton.

If I compute the expectation of this I get f representing the net profit:

f(x)= 59.50 * n -25 - 200 E(X^2) = 59.50n-25-200 (0.05n)^2
(since X follows a binomial distribution with p = 0.05)

Computing the derivative of this function ( to find maximum point) I get n = 59.50. However, the answer is 50. I am not sure what I failed to consider in this question.

2. ## Re: Net profit Expectation Question

59.50n-25-200 (0.05n)^2
The stuff in red is $\displaystyle \left(E[X] \right)^2$, it should be $\displaystyle E[X^2]$

Correct that and see if you end up with the right answer. If you dont know $\displaystyle E[X^2]$ you can calculate it using: $\displaystyle E[X^2] = Var(X) + \left(E[X] \right)^2$