he average cost per item is C(x) / x

(2x^3 – 4x^2 + 24000x) / x =>

2x^2 – 4x + 24000

A(x) = 2x^2 – 4x + 24000

A'(x) = 4x – 4

A'(x) = 0

0 = 4x – 4

0 = x – 1

x = 1

So, when x = 1, the average cost per item is at its lowest amount.

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# C(x) = 2x^3 – 4x^2 + 24,000x is the cost of manufacturing x items ,Suppose that

Science & Mathematics

he average cost per item is C(x) / x

(2x^3 – 4x^2 + 24000x) / x =>

2x^2 – 4x + 24000

A(x) = 2x^2 – 4x + 24000

A'(x) = 4x – 4

A'(x) = 0

0 = 4x – 4

0 = x – 1

x = 1

So, when x = 1, the average cost per item is at its lowest amount.