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The Total Cost of Producing X Radio Sets per Day is Rs ( X 2 4 + 35 X + 25 ) and the Price per Set at Which They May Be Sold is Rs. ( 50 − X 2 ) . Ind the - CBSE (Science) Class 12 - Mathematics

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Question

The total cost of producing x radio sets per  day is Rs \[\left( \frac{x^2}{4} + 35x + 25 \right)\] and the price per set  at which they may be sold is Rs. \[\left( 50 - \frac{x}{2} \right) .\] Find the daily output to maximum the total profit.

Solution

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

\[ \Rightarrow P = x\left( 50 - \frac{x}{2} \right) - \left( \frac{x^2}{4} + 35x + 25 \right)\]

\[ \Rightarrow P = 50x - \frac{x^2}{2} - \frac{x^2}{4} - 35x - 25\]

\[ \Rightarrow \frac{dP}{dx} = 50 - x - \frac{x}{2} - 35\]

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

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

\[ \Rightarrow 15 - \frac{3x}{2} = 0\]

\[ \Rightarrow 15 = \frac{3x}{2}\]

\[ \Rightarrow x = \frac{30}{3}\]

\[ \Rightarrow x = 10\]

\[\text { Now,} \]

\[\frac{d^2 P}{d x^2} = \frac{- 3}{2} < 0\]

\[\text{ So, profit is maximum if daily output is 10 items.}\]

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Solution The Total Cost of Producing X Radio Sets per Day is Rs ( X 2 4 + 35 X + 25 ) and the Price per Set at Which They May Be Sold is Rs. ( 50 − X 2 ) . Ind the Concept: Graph of Maxima and Minima.
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