Manufacturer can sell x items at a price of rupees


Manufacturer can sell $x$ items at a price of rupees $\left(5-\frac{x}{100}\right)$ each. The cost price is Rs $\left(\frac{x}{5}+500\right) .$ Find the number of items he should sell to earn maximum profit.


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

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

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

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

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

$\Rightarrow x=240$


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

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.

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