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प्रश्न
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 revenue is increasing
उत्तर
Let C be the total cost function and R be the revenue
∴ C = 180 + 4x
Now, Revenue = Price × Demand
∴ R = P × x = (240 – x)x
∴ R = 240x – x2
∴ `"dR"/("d"x)` = 240 – 2x
= 2(120 – x)
Since revenue R is an increasing function, `"dR"/("d"x)` > 0
∴ 2(120 – x) > 0
∴ 120 – x > 0
∴ 120 > x
∴ x < 120
∴ The revenue is increasing for x < 120.
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संबंधित प्रश्न
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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 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 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`
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Ec = (0.0003)I2 + (0.075)I2
when I = 1000
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