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Question
A manufacturing company produces x items at the total cost of Rs (180 + 4x). The demand function of this product is P = (240 − x). Find x for which profit is increasing.
Solution
Total cost function (C) = 180 + 4x
Demand function (P) = 240 − x
Where x is the number of items produced.
Total revenue (R) = P × D
∴ R = x (240 − x)
∴ R = 240x − x2
Profit function π = R − C
∴ π = (240x − x2) − (180 + 4x)
∴ π = 240x − x2 − 180 − 4x
∴ π = − x2 + 236x − 180
Differentiating w.r.t.x,
∴ `"dπ"/"dx"` = − 2x + 236
Profit π is increasing if `"dπ"/"dx"` > 0
i.e. if − 2x + 236 > 0
i.e. if 236 > 2x
i.e. if x < `236/2`
i.e. if x < 118
∴ The profit is increasing for x < 118.
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The manufacturing company produces x items at the total cost of ₹ 180 + 4x. The demand function for this product is P = (240 − 𝑥). Find x for which profit is increasing
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`
