# The cost (in dollars) of producing x units of a certain commodity is C(x) = 8000 + 6x + 0.15x2. (a) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to the given value. (Round your answers to the nearest cent.) (i) x = 101 \$__________per unit. (b) Find the instantaneous rate of change of C with respect to x when x = 100. (This is called the marginal cost.) \$____________per unit. a) You need to remember that the average rate of change is the change in y value over the change in x value for two distinct points, hence if x changes from 10 to 101, then you may evaluate the average rate of change of `C(x)`  such that:

`(C(101) - C(100))/(101...

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a) You need to remember that the average rate of change is the change in y value over the change in x value for two distinct points, hence if x changes from 10 to 101, then you may evaluate the average rate of change of `C(x)`  such that:

`(C(101) - C(100))/(101 - 100)`

You need to evaluate `C(101)`  such that:

`C(101) = 8000 + 0.15*101^2 = 9530.15`

You need to evaluate `C(100)`  such that:

`C(100) = 8000 + 0.15*100^2 = 9500`

`(C(101) - C(100))/(101 - 100) = (9530.15 - 9500)/1`

`(C(101) - C(100))/(101 - 100) = 30.15`

Hence, evaluating the average rate of change of C(x) considering the given changes yields `(C(101) - C(100))/(101 - 100) = 30.15.`

b) You need to find the instantaneous rate of change, hence, you need to differentiate the function C(x) with respect to x such that:

`C'(x) = 0.3x`

You need to evaluate C'(x) at x = 100 such that:

`C'(100) = 0.3*100 = 30`

Hence, evaluating the instantaneous rate of change at `x = 100`  yields `C'(100) = 30.`

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