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The profit function P(x) of a firm, selling x items per day is given by P(x) = (150 – x)x – 1625 . Find the number of items the firm should manufacture to get maximum profit. Find the maximum profit.

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प्रश्न

The profit function P(x) of a firm, selling x items per day is given by P(x) = (150 – x)x – 1625 . Find the number of items the firm should manufacture to get maximum profit. Find the maximum profit.

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उत्तर

Profit function P(x) is given by

P(x) = (150 – x)x – 1625

= 150x – x2 – 1625

∴ P'(x) = `"d"/("d"x)(150 x - x^2 - 1625)`

= 150 × 1 – 2x – 0

= 150 – 2x

and

P"(x) = `"d"/("d"x)(150 - 2x)`

= 0 – 2 × 1

= – 2

Now, P'(x) = 0 gives, 150 – 2x = 0

∴ x = 75

and

P''(75) = – 2 < 0

∴ by the second derivative test, P(x) is maximum when x = 75

Maximum profit = P(75)

= (150 – 75)75 – 1625

= 75 × 75 – 1625

= 4000

Hence, the profit will be maximum, if the manufacturer manufactures 75 items and maximum profit is 4000.

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अध्याय 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ९०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 2 Applications of Derivatives
Exercise 2.4 | Q 18 | पृष्ठ ९०

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