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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.

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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.

 

Sum
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Solution

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

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

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

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

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

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

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

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

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

\[ \Rightarrow x = 240\]

\[\text { Now, }\]

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

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

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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.

  Is there an error in this question or solution?
Chapter 17: Maxima and Minima - Exercise 18.5 [Page 74]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 17 Maxima and Minima
Exercise 18.5 | Q 37 | Page 74
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