Indian Institute of Foreign Trade 2010 Solved Paper

It is given that the fare charged (in RS.) is defined by the function $\mathit{f}={\left(54-\frac{\mathit{x}}{32}\right)}^2$ where 'x' is the number of the people per trip
Let 'x' be the number of people per trip that will make the marginal revenue equal to zero.
Total revenue generated per trip, $\mathit{g}\left(\mathit{x}\right)=\mathit{x}\times {\left(54-\frac{\mathit{x}}{32}\right)}^2$
When total revenue is maximum, then the marginal revenue will be zero. Therefore, we have to find out the value of 'x' for which $\mathit{g}\prime \left(\mathit{x}\right)=0$
$\Rightarrow {\left(54-\frac{\mathit{x}}{32}\right)}^2-\frac{2\mathit{x}}{32}\times \left(54-\frac{\mathit{x}}{32}\right)=0$
$\Rightarrow {\left(54-\frac{\mathit{x}}{32}\right)}^2-\frac{2\mathit{x}}{32}\times \left(54-\frac{\mathit{x}}{32}\right)=0$
$\Rightarrow \left(54-\frac{\mathit{x}}{32}\right)\times \left(54-\frac{3\mathit{x}}{32}\right)=0$
$\Rightarrow \mathit{x}=1728,576$
It is given that the train has a capacity of 900 only. Hence $\mathit{x}\ne 1728$ i.e. $\mathit{x}=576$
Therefore, option B is the correct answer.