Question

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.

Answer 1

$$\begin{gathered} \text{Profit =S.P. - C.P.} \\ \Rightarrow P=x\left(5-\frac{x}{100}\right)-\left(500+\frac{x}{5}\right) \\ \Rightarrow P=5x-\frac{x^2}{100}-500-\frac{x}{5} \\ \Rightarrow \frac{dP}{dx}=5-\frac{x}{50}-\frac{1}{5} \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}P,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dP}{dx}=0 \\ \Rightarrow 5-\frac{x}{50}-\frac{1}{5}=0 \\ \Rightarrow \frac{24}{5}=\frac{x}{50} \\ \Rightarrow x=\frac{24\times 50}{5} \\ \Rightarrow x=240 \\ \mathrm{Now}, \\ \frac{d^{2}P}{d{x}^2}=\frac{-1}{50}<0 \\ \text{So, the profit is maximum if 240 items are sold.} \end{gathered}$$

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