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Manufacturer Can Sell X Items at a Price of Rupees ( 5 − X 100 ) Each. the Cost Price is Rs ( X 5 + 500 ) . Find the Number of Items He Should Sell to Earn Maximum Profit. - Mathematics

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प्रश्न

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.

 

योग

उत्तर

\[\text { Profit =S.P. - C.P.}\]

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

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

\[ \Rightarrow \frac{dP}{dx} = 5 - \frac{x}{50} - \frac{1}{5}\]

\[\text { For maximum or minimum values of P, we must have }\]

\[\frac{dP}{dx} = 0\]

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

\[ \Rightarrow \frac{24}{5} = \frac{x}{50}\]

\[ \Rightarrow x = \frac{24 \times 50}{5}\]

\[ \Rightarrow x = 240\]

\[\text { Now, }\]

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

\[\text { So, the profit is maximum if 240 items are sold.}\]

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Notes

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

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Maxima and Minima - Exercise 18.5 [पृष्ठ ७४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 18 Maxima and Minima
Exercise 18.5 | Q 37 | पृष्ठ ७४

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