Manufacturer can sell x items at a price of rupees

Profit =S.P. - C.P.

$\Rightarrow P=x\left(5-\frac{x}{100}\right)-\left(500+\frac{x}{5}\right)$

$\Rightarrow P=5 x-\frac{x^{2}}{100}-500-\frac{x}{5}$

$\Rightarrow \frac{d P}{d x}=5-\frac{x}{50}-\frac{1}{5}$

For maximum or minimum values of $P$, we must have

$\frac{d P}{d x}=0$

$\Rightarrow 5-\frac{x}{50}-\frac{1}{5}=0$

$\Rightarrow \frac{24}{5}=\frac{x}{50}$

$\Rightarrow x=\frac{24 \times 50}{5}$

$\Rightarrow x=240$

Now,

$\frac{d^{2} P}{d x^{2}}=\frac{-1}{50}<0$

So, the profit is maximum if 240 items are sold.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.