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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.
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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 − 2x = 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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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`
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 elasticity of demand for price ₹ 80.
Solution: Total cost C = 40 + 2x and Price p = 120 – x
p = 120 – x
∴ x = 120 – p
Differentiating w.r.t. p,
`("d"x)/("dp")` = `square`
∴ Elasticity of demand is given by η = `- "P"/x*("d"x)/("dp")`
∴ η = `square`
When p = 80, then elasticity of demand η = `square`
If elasticity of demand η = 0 then demand is ______.
If f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.
