Business Application

Calculus
Section 2.7



We end Section 2.7 with this problem.

I would like to see you work out part (a) (i) only.
I also would like to see how (b) is done.

My weekend is over.
Sad but true.

 

C(x)=5000+10x+0.05x^2

a. find the avarage rate of change of C with respect to x when the production level is changed

(i) from x=100 to x=105

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 100 to 105, then you may evaluate the average rate of change of

C(x) such that: (C(105)-C(100))/(105 - 100)

C(105) =6601.25
C(100) =6500

the average rate of change is=(6601.25-6500)/(105 - 100)=20.25
.

b) find the instantaneous rate of change of C with respect to x when x=100

The instantaneous rate of change of C with respect to x=100 is C'(x)

C'(x) =10+2*0.05x
C'(x) =10+0.1x .......if x=100
C'(x) =10+0.1*100
C'(x) =20

 

We take the derivative of the original function C(x). Yes? If this is true, where did C '(x) =10+2*0.05x come from?

 

C'(x) is derivative of the original function C(x)

 

Understood.