calculate price based on demands and maximize revenue I believe I have a simple question which I am struggling to answer. It is as follows:
We have 400 items, each item costs £100. Retailer bought these items before the season started. 
The forecasted demands for each of the three months (length of the season) are:
$d_{1} = 300 - p_{1}$, $d_{2} = 300 - 1.3p_{2}$, $d_{3} = 300 - 1.8p_{3}$, $d_{i}$  are the demands each months and $p_{i}$ are the prices respectively.
The questions are:
Q1. If the retailer charges a constant price over the three months, what will it be? 
I attempted to answer this part and I got p = £121.95. Could someone verify whether this is correct?
Q2. How should retailer vary the price of the item over the three months in order to maximize the revenue?
Any chances someone could help me out on both of these questions?
 A: It's not clear to me what you mean by "each item costs £100", since the price is yet to be determined – I'll just ignore that.
I'll write the individual demands as
$$
d_i=300-\mu_ip_i\;.
$$
Question 1: If the retailer charges a constant price $p$, the total demand will be 
$$
\sum_i\left(300-\mu_ip\right)=900-4.1p\;,
$$
so the retailer will sell $\min(900-4.1p,400)$ items, earning $\min(900-4.1p,400)\cdot p$. We have $900-4.1p=400$ at $p=500/4.1$, so that's the maximal price at which the retailer can sell all $400$ items, whereas by increasing the price beyond that point she can earn $(900-4.1p)p$, but this is maximal for $p=900/8.2\lt500/4.1$, so the optimal price is $500/4.1\approx121.95$, as you found.
Question 2: The total demand must be $\le400$, since if it's greater we could earn more by increasing one of the prices. So we can optimise without constraints on the prices, and if the total demand we find exceeds $400$, the solution must have total demand $400$. Optimizing the revenue from selling $300-\mu p$ items at price $p$ yields $p=150/\mu$ and thus a demand of $150$, so the total demand of the unconstrained problem is $450\gt400$. Thus we need to optimise under the constraint
$$
\sum_i\left(300-\mu_ip_i\right)=400\;.
$$
With a Lagrange multiplier, this yields the objective function
$$
\sum_i\left(300-\mu_ip_i\right)p_i+\lambda\sum_i\mu_ip_i\;,
$$
setting the derivative with respect to $p_i$ to $0$ yields
$$
300-2\mu_ip_i+\lambda\mu_i=0
$$
and thus
$$
p_i=\frac{300}{2\mu_i}+\frac\lambda2\;,
$$
and substituting into the constraint then yields
$$
\frac\lambda2\sum_i\mu_i=50\;,
$$
and thus
$$
\lambda=\frac{100}{4.1}\;.
$$
Thus the optimal prices are
\begin{align}
p_1=\frac{300}2+\frac{50}{4.1}\approx162.20\;,\\
p_2=\frac{300}{2\cdot1.3}+\frac{50}{4.1}\approx127.58\;,\\
p_3=\frac{300}{2\cdot1.8}+\frac{50}{4.1}\approx67.75\;,
\end{align}
for a total revenue of
$$
\sum_i\left(300-\mu_ip_i\right)p_i=150\sum_ip_i\approx51697.94\;,
$$
compared to $400\cdot500/4.1\approx48780.49$ with the constant price.
