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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 :

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

[tex]A=\frac{C(105)-C(100)}{105-100}[/tex][tex]A=\frac{(5500+8(105)+0.6(105)^2)-(5500+8(100)+0.6(100)^2)}{5}[/tex][tex]A=\frac{(5500+840+6615)-(5500+800+6000)}{5}[/tex][tex]A=\frac{12955-12300}{5}[/tex][tex]A=\frac{655}{5}=131[/tex]

(in) from x = 100 to x = 101

[tex]A=\frac{C(101)-C(100)}{101-100}[/tex][tex]A=\frac{(5500+8(101)+0.6(101)^2)-(5500+8(100)+0.6(100)^2)}{1}[/tex][tex]A=(5500+808+6120.6)-(5500+800+6000)[/tex][tex]A=12428.6-12300[/tex][tex]A=128.6[/tex]

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

[tex]C´(x)=1.2x+8[/tex]

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

[tex]C^{\prime}(100)=1.2(100)+8[/tex][tex]C^{\prime}(100)=120+8[/tex][tex]C^{\prime}(100)=128[/tex]

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