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प्रश्न
The total cost of manufacturing x articles C = 47x + 300x2 – x4 . Find x, for which average cost is decreasing
उत्तर
Given, total cost function is
C = 47x + 300x2 – x4
Average cost CA = `"C"/"A"`
∴ CA = `(47 x + 300x^2 - x^4)/x`
= `(x(47 + 300x - x^3))/x`
∴ CA = 47 + 300x – x3
∴ `"dC"_"A"/"dx"` = 0 + 300x – 3x3
= 300x – 3x3
= 3(100 – x2)
Since average cost CA is a decreasing function `"dC"_"A"/"dx" < 0`
∴ 3(100 – x2) < 0
∴ 100 – x2 < 0
∴ 100 < x2
∴ x2 > 100
∴ x > 10 or x < – 10
But x < – 10 is not possible .....[∵ x > 0]
∴ x > 10
∴ The average cost CA is decreasing for x > 10.
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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 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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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`
