मराठी

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.

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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.

 

बेरीज
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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.

  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 17: Maxima and Minima - Exercise 18.5 [पृष्ठ ७४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 17 Maxima and Minima
Exercise 18.5 | Q 37 | पृष्ठ ७४
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