cost per unit = cpu = C(x) / x
= 36,100/x + 100 +.01 x
d cpu/dx = 0 at min
= -36,100/x^2 + .01
.01 x^2 = 36,100
x^2 = 3,610,000
x = 1900 units
now go back and get cpu
= 36,100/1900 + 100 + .01(1900)
= 2019
Assume that it costs Apple approximately
C(x) = 36,100 + 100x + 0.01x^2
dollars to manufacture x 30-gigabyte video iPods in a day.
How many iPods should be manufactured in order to minimize average cost?
What is the resulting average cost of an iPod? (Give your answer to the nearest dollar.)
2 answers
d cpu/dx = 0 at min
= -36,100/x^2 + .01
.01 x^2 = 36,100
x^2 = 3,610,000
x = 1900 units
now go back and get cpu
= 36,100/1900 + 100 + .01(1900)
What is the resulting average cost of an iPod?
138
= -36,100/x^2 + .01
.01 x^2 = 36,100
x^2 = 3,610,000
x = 1900 units
now go back and get cpu
= 36,100/1900 + 100 + .01(1900)
What is the resulting average cost of an iPod?
138