Question

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.

Answer 1

Answer:

(a)$C'(x)=\dfrac{x^2-2560000}{x^2}$

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

(c)$C''(x)=\dfrac{5120000}{x^3}$

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:

$C(x)=25,600+100x+0.01x^2$

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

$Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}$

The average cost per iPod is therefore:

$C'(x)=\dfrac{x^2-2560000}{x^2}$

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

$C'(x)=\dfrac{x^2-2560000}{x^2}=0\\x^2-2560000=0\\x^2=2560000\\x=\sqrt{2560000}=1600$

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

$Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}\\\dfrac{25,600+100(1600)+0.01(1600)^2}{1600}\\=\ 132$

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.

$C''(x)=\dfrac{5120000}{x^3}$

At the critical point,  x=1600

$C''(1600)=\dfrac{5120000}{1600^3}=0.00125$

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.