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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 revenue is increasing.
Solution
Let C be the total cost function.
∴ C = 40 + 2x
Revenue = Price × Demand
∴ `"R" = "p" × "x" = (120 - "x") * "x"`
∴ R = 120x - x2
∴ `"dR"/"dx" = 120 - 2"x" = 2(60 - "x")`
Since revenue R is an increasing function, `"dR"/"dx" > 0`
∴ 2(60 - x) > 0
∴ 60 - x > 0
∴ 60 > x
∴ x < 60
∴ The revenue R is increasing for x < 60.
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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
Solution: Total cost C = 40 + 2x and Price p = 120 – x
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`
If elasticity of demand η = 0 then demand is ______.
If f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.
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`
