Question

A manfufacture can sell $x$ items at a price of $Rs(5-\frac{x}{100})$ each. The cost price of $x$ items is $Rs(\frac{x}{5}+500).$ The number of items he should sell to earn maximum profit is equal to

Answer 1

Selling price of $x$ items $=(5-\frac{x}{100})x$
Cost price of $x$ items $=\frac{x}{5}+500$
$\therefore$ Profit, $P\left(x\right)=$ Selling price $-$ Cost price
$=(5-\frac{x}{100})x-(\frac{x}{5}+500)$
$P\left(x\right)=\frac{24}{5}x-\frac{x^2}{100}-500$
$\Rightarrow P^{\prime }\left(x\right)=\frac{24}{5}-\frac{x}{50}$ and $P^{\prime \prime }\left(x\right)=-\frac{1}{50}<0$
$\therefore P\left(x\right)$ is maximum when $P^{\prime }\left(x\right)=0$
$\Rightarrow x=240$