The given cost function is:

C(x)=x3+34x+250C(x)=x3+34x+250

Where, x is the number of units.

The average cost is the total cost divided by the number of units:

A(x)=C(x)x=x3+34x+250xA(x)=x2+34+250xA(x)=C(x)x=x3+34x+250xA(x)=x2+34+250x

Differentiating with respect to x, we get:

A′(x)=2x−250x2A′(x)=02x−250x2=02x3−250=0x3=125x=5A′(x)=2x−250x2A′(x)=02x−250x2=02x3−250=0x3=125x=5

This is the critical point for A(x).

(a)

We have,

1≤x≤101≤x≤10

x = 5 lies in the given interval.

At x = 1

A(1)=12+34+2501A(1)=285A(1)=12+34+2501A(1)=285

At x = 5

A(5)=52+34+2505A(5)=109A(5)=52+34+2505A(5)=109

At x = 10

A(10)=102+34+25010A(10)=159A(10)=102+34+25010A(10)=159

Thus, in this interval the minimum average cost is, 109.

(b)

We have,

10≤x≤2010≤x≤20

x = 5 does not lies in the given interval.

At x = 10

A(10)=102+34+25010A(10)=159A(10)=102+34+25010A(10)=159

At x = 20

A(20)=202+34+25020A(20)=446.5A(20)=202+34+25020A(20)=446.5

Thus, in this interval the minimum average cost is, 159.