Question

The cost (in dollars) of producing a units of a certain commodity is C(x) = 5500 + 8x + 0.6x^2(a) Find the average rate of change of C' with respect to a when the production level is changed(i) from x = 100 to x = 105.Average rate of change =(in) from x = 100 to x = 101.Average rate of change =(b) Find the instantaneous rate of change of C with respect to a when x = 100. (This is called the marginal cost.)Instantaneous rate of change =

Answer 1

The average rate of change is calculated dividing the change of C(x) by the change of x,

(i) from x = 100 to x = 105

`A=(C(105)-C(100))/(105-100)`
`A=((5500+8(105)+0.6(105)^2)-(5500+8(100)+0.6(100)^2))/(5)`
`A=((5500+840+6615)-(5500+800+6000))/(5)`
`A=(12955-12300)/(5)`
`A=(655)/(5)=131`

(in) from x = 100 to x = 101

`A=(C(101)-C(100))/(101-100)`
`A=((5500+8(101)+0.6(101)^2)-(5500+8(100)+0.6(100)^2))/(1)`
`A=(5500+808+6120.6)-(5500+800+6000)`
`A=12428.6-12300`
`A=128.6`

To find the instantaneous rate of change we need to find the derivative of C(x),

`C´(x)=1.2x+8`

(b) Find the instantaneous rate of change of C with respect to a when x = 100.

`C^(\prime)(100)=1.2(100)+8`
`C^(\prime)(100)=120+8`
`C^(\prime)(100)=128`