Question

A company rents out computer time for periods of t hours for which it receives rs. 600 per hour. The number of times the computer breaks down during t hours is a random variable having the poisson distribution with mean 0.8 .T if the computer breaks down x times during t hours, it costs 2 50x rupees to fix it. How should the company select t in order to maximize its expected profit?

Answer 1

Answer:

i do not know sorry

i am interested in the medical

Answer 2

Answer:

ExplanatLet's denote the number of breakdowns during $t$ hours by $X$, then $X$ follows a Poisson distribution with mean $\lambda=0.8t$. The profit generated by renting out the computer for $t$ hours is $600t$. The expected cost of repairing the computer is $E[250X]=250E[X]=200t$, since the cost of fixing a breakdown is 250 rupees and the expected number of breakdowns is $\lambda$.

Thus, the expected profit can be expressed as:

$P(t)=600t-200t=400t$

To maximize this expression, we need to find the value of $t$ that maximizes $P(t)$. Taking the derivative of $P(t)$ with respect to $t$, we get:

$P'(t) = 400$

Since the derivative is positive for all $t$, the function $P(t)$ is increasing for all $t$. Therefore, to maximize the expected profit, the company should select the largest possible value of $t$.

However, we also need to consider the fact that customers may be less likely to rent the computer for longer periods, and therefore we may need to balance the expected profit with the likelihood of customers renting the computer. If this is a concern, the company could consider using a more complex model that takes into account the probability of customers renting the computer for different lengths of timeion: