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Question
A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which profit is increasing.
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Solution
Let C be the total cost function.
∴ C = 40 + 2x
Profit = Revenue - Cost
∴ π = R - C
∴ π = 120x - x2 - (40 + 2x)
= 120x - x2 - 40 - 2x
∴ π = `- "x"^2 + 118"x" - 40`
∴ `("d"pi)/"dx" = - 2"x" + 118 = 2(- "x" + 59)`
Since profit π is an increasing function, `("d"pi)/"dx" > 0`
∴ 2(- x + 59) > 0
∴ - x + 59 > 0
∴ 59 > x
∴ x < 59
∴ The profit π is increasing for x < 59.
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A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price per item is given as p = 120 – x. Find the value of x for which revenue is increasing
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Revenue R = `square`
Differentiating w.r.t. x,
∴ `("dR")/("d"x) = square`
Since Revenue is increasing,
∴ `("dR")/("d"x)` > 0
∴ Revenue is increasing for `square`
A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price per item is given as p = 120 – x. Find the value of x for which profit is increasing
Solution: Total cost C = 40 + 2x and Price p = 120 − x
Profit π = R – C
∴ π = `square`
Differentiating w.r.t. x,
`("d"pi)/("d"x)` = `square`
Since Profit is increasing,
`("d"pi)/("d"x)` > 0
∴ Profit is increasing for `square`
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Solution: Total cost C = 40 + 2x and Price p = 120 – x
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∴ x = 120 – p
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