Management Aptitude Test Sep 2020 Paper

Let the carpenter increases the price $\mathit{x}$ times, then the total increase in price would be ₹ $6\mathit{x}$. Also, the chair sold at increased price $=\left(40-\mathit{x}\right)$ To maximize the profit the selling price of chair with increase price must be greater than original selling price,
$\therefore \left(156+6\mathit{x}\right)\left(40-\mathit{x}\right)>156\times 40$
$156\times 40+240\mathit{x}-156\mathit{x}-6{\mathit{x}}^2>156\times 40$
$84\mathit{x}-6{\mathit{x}}^2>0$
Differentiate the above equation and putting = 0
$84-12\mathit{x}=0$
$\Rightarrow 84=12\mathit{x}$
$\therefore \mathit{x}=\frac{84}{12}=7$
To check, if value of $\mathit{x}$ will maximize or not we will against differentiate the above equation and put the value of $\mathit{x}$ and if the result is negative, then the profit is maximized and if positive, then profit is minimum
So, on differentiating $84-12\mathit{x}=-12$
So, the result is negative, hence the profit is maximum.
The maximum selling price $=156+6\times 7$