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Question
The cost function of a commodity is `C(x) = 200 + 20x - 1/2x^2` (in rupees). Find the range where AC falls.
Solution
`C(x) = 200 + 20x - 1/2x^2`
`\implies AC = 200/x + 20 - 1/2x`
To find AC falls:
`d/dx(AC) = d/dx(200/x + 20 - 1/2x)`
= `-200/x^2 - 1/2`
= `-(200/x^2 + 1/2) < 0, ∀ x > 0`
Hence AC falls continuously.
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