A manufacturer sells $x$ items at price $\left(5-\dfrac{x}{100}\right)$ each; cost of $x$ items is $\left(\dfrac{x}{5}+500\right)$. The profit is maximum when the number of items sold is

1. $P(x)=\left(5-\dfrac{x}{100}\right)x-\left(\dfrac{x}{5}+500\right)=\dfrac{24}{5}x-\dfrac{x^2}{100}-500$. 2. $P'(x)=\dfrac{24}{5}-\dfrac{x}{50}$. 3. $P'(x)=0 \Rightarrow x=50\cdot\dfrac{24}{5}=240$. 4. $P''(x)=-\dfrac{1}{50}<0$, so $x=240$ is a maximum. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.36_