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Solution for 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. - CBSE (Commerce) Class 12 - Mathematics

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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.

 

Solution

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

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

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

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

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

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

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

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

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

\[ \Rightarrow x = 240\]

\[\text { Now, }\]

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

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

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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Solution for question: 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. concept: Graph of Maxima and Minima. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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