Assume that it costs Apple approximately E(x) 25,600 + 100x + 0.012 dollars to manufacture x 32GB iPods in a day. (a) The average cost per iPod when they manufacture x iPods in a day is given by C(x)- (b) How many iPods should be manufactured in order to minimize average cost? iPods per day What is the resulting minimum average cost of an iPod? (Give your answer to the nearest dollar.) dollars Second derivative test: Your answer above is a critical point for the average cost function. To show it is a minimum, calculate the second derivative of the average cost function. C"(x) Evaluate C(x) at your critical point. The result is -Select. , which means that the average cost is Selct. at the critical point, and the critical point is a minimum.

(a)

(b)x=1600, Minimum Average Cost Per iPod=$132

(c)

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

Step-by-step explanation:

Given that it costs Apple approximately $ C(x) to manufacture x 32GB iPods in a day, where:

(a)The average cost per iPod when they manufacture x iPods in a day is given by:

The average cost per iPod is therefore:

(b)To minimize average cost of x iPods per day, we set the average cost per iPod=0 and solve for x.

The resulting minimum average cost (at x=1600) is given as:

Second derivative test

(c)The answer above is a critical point for the average cost function. To show it is a minimum, we calculate the second derivative of the average cost function.

At the critical point,  x=1600

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

4th option is correct xy=-10

if this was to u plzz mark me the brainliest

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).