If c(x) is the total cost incurred in producing x units of a certain commodity, then the average cost of producing x units of the commodity is given by c¯ = c(x) x , for x > 0. the total monthly cost (in dollars) incurred by cannon precision instruments for producing x units of the model m1 camera is given by c(x) = 0.0025x 2 + 80x + 10, 000. how many cameras should cannon produce in a month in order to have the lowest possible average monthly cost?

x  =  2000 cameras

Step-by-step explanation:

C(x)     Total cost in producing  x units

C- = C(x) /x   Average cost of producing x units    x > 0

Cannon Precision Instrument

C (x)   Total monthly cost for producing x units of M1 cameras

is    C(x)  =  0.0025x²  + 80x  +  10000

Then average cost of producing x cameras M1 is

C-(x)  =  ( 0.0025x²  + 80x  +  10000)  /x

C-(x)  =   0.0025x  + 80  + 10000/x

Taking derivatives on both sides of the equation

C-´(x)   =  0.0025   -  10000/x²

Then

C-´(x)   = 0

( 0.0025x² - 10000 ) / x²   =  0

0.0025x² - 10000 =  0

x²  =  10000 /0.0025         x²  = 4000000

x  =  2000 cameras

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !