Manufacturer can sell x items at a price of rupees ≤ft( 5 - x/100 ) e…

The correct answer is Option (2) → 240 items ##

Let $S(x)$ be the selling price of $x$ items and let $C(x)$ be the cost price of $x$ items. Then, we have

$S(x) = \left( 5 - \frac{x}{100} \right)x = 5x - \frac{x^2}{100}$

and $C(x) = \frac{x}{5} + 500$

Thus, the profit function $P(x)$ is given by

$P(x) = S(x) - C(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500$

i.e. $P(x) = \frac{24}{5}x - \frac{x^2}{100} - 500$

or $P'(x) = \frac{24}{5} - \frac{x}{50}$

Now $P'(x) = 0$ gives $x = 240$. Also $P''(x) = \frac{-1}{50}$. So $P''(240) = \frac{-1}{50} < 0$

Thus, $x = 240$ is a point of maxima. Hence, the manufacturer can earn maximum profit, if he sells 240 items.