Find the number of units x that produces the minimum average cost per unit C in the given equation. C = 0.001x³ + 5x + 250

Solution:

It is given that,

C = 0.001x³ + 5x + 250

Then,

unit cost f(x) = C/x

= 0.001x³/x + 5x/x+ 250/x

So we get

f(x) = 0.001x² + 5 + 250/x

To find the minimum cost, let us differentiate f(x).

f'(x) = 0.002x - 250/x²

Now we have to equate the first derivative to zero

0.002x - 250/x² = 0

0.002x = 250/x²

By cross multiplication we get,

0.002x × x² = 250

0.002x³ = 250

Divide both sides by 0.002

x³ = 250/0.002

x³ = 125000

So we get

x = √(125000)

x = 50 units

Therefore, the number of units x that produces the minimum average cost per unit C is 50 units.

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Find the number of units x that produces the minimum average cost per unit C in the given equation. C = 0.001x³ + 5x + 250

Summary:

The number of units x that produces the minimum average cost per unit C in the given equation C = 0.001x³ + 5x + 250 is 50 units.