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Cost of Assembling X Wallclocks is X^3/3 - 40x^2) and Labour Charges Are 500x. Find the Number of Wallclocks to Be Manufactured for Which Marginal Cost is Minimum, - Mathematics and Statistics

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Sum

Cost of assembling x wallclocks is `( x^3/3 - 40x^2)` and labour charges are 500x. Find the number of wall clocks to be manufactured for which average cost and marginal cost attain their respective minimum.

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Solution

Total cost = Assembling cost + Labour cost
Total cost = `x^3/3 - 40x^2 + 500x`
C(x) = `x^3/3 - 40x^2 + 500x`

`C_M = (dc)/(dx)`
`C_M = x^2 - 80x + 500`

`(dC_x)/dx = 2x - 80`

`C_M  "is minimum if  "(dC_M)/dx = 0 and (d^2C_M)/(dx^2) > 0`

∴ `(d^2C_M)/dx = 0`

∴ 2x - 80 = 0

∴ x = 40

`(d^2C_M)/(dx^2) = 2`

`(d^2C_M)/(dx^2)` > 0 at x = 40

∴ Marginal cost is minimum at x = 40.

Concept: Maxima and Minima
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