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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.
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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In a factory, for production of Q articles, standing charges are ₹500, labour charges are ₹700 and processing charges are 50Q. The price of an article is 1700 - 3Q. Complete the following activity to find the values of Q for which the profit is increasing.
Solution: Let C be the cost of production of Q articles.
Then C = standing charges + labour charges + processing charges
∴ C = `square`
Revenue R = P·Q = (1700 - 3Q)Q = 1700Q- 3Q2
Profit `pi = R - C = square`
Differentiating w.r.t. Q, we get
`(dpi)/(dQ) = square`
If profit is increasing , then `(dpi)/(dQ) >0`
∴ `Q < square`
Hence, profit is increasing for `Q < square`
