If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost function C(x) given below.

C(x) = 54,000 + 240x + 4x^3/2

Required:
a. Find the total cost at a production level of 1000 units.
b. Find the average cost at a production level of 1000 units.
c. Find the marginal cost at a production level of 1000 units.
d. Find the production level that will minimize the average cost.
e. What is the minimum average cost?

Respuesta :

Answer:

a. C(1,000) = 420,491.11

b. c(1,000) = 420.49

c. dC/dx(1,000) = 429.72

d. x = 900

e. c(900) = 420

Step-by-step explanation:

We have a cost function for x units written as:

[tex]C(x) = 54,000 + 240x + 4x^{3/2}[/tex]

a. The total cost for x=1000 units is:

[tex]C(1,000) = 54,000 + 240(1,000) + 4(1,000)^{3/2}\\\\C(1,000)=54,000+240,000+4\cdot 31,622.78\\\\C(1,000)=54,000+240,000+ 126,491.11 \\\\C(1,000)= 420,491.11[/tex]

b. The average cost c(x) can be calculated dividing the total cost by the amount of units:

[tex]c(1,000)=\dfrac{C(1,000)}{1,000}=\dfrac{ 420,491.11 }{1,000}= 420.49[/tex]

c. The marginal cost can be calculated as the first derivative of the cost function:

[tex]\dfrac{dC}{dx}=240(1)+4(3/2)x^{3/2-1}=240+6x^{1/2}\\\\\\\dfrac{dC}{dx}(1,000)=240+6(1,000)^{1/2}=240+6\cdot 31.62=429.72[/tex]

d. This value for x, that minimizes the average cost, happens when the first derivative of the average cost is equal to 0.

[tex]c(x)=\dfrac{C(x)}{x}=\dfrac{54,000+240x+4x^{3/2}}{x}=54,000x^{-1}+240+4x^{1/2}\\\\\\ \dfrac{dc}{dx}=54,000(-1)x^{-2}+0+4(1/2)x^{-1/2}=0\\\\\\\dfrac{dc}{dx}=-54,000x^{-2}+2x^{-1/2}=0\\\\\\2x^{-1/2}=54,000x^{-2}\\\\\\x^{-1/2+2}=54,000/2=27,000\\\\\\x^{3/2}=27,000\\\\\\x=27,000^{2/3}=900[/tex]

e. The minimum average cost is:

[tex]c(900)=54,000(900)^{-1}+240+4(900)^{1/2}\\\\c(900)=60+240+120\\\\c(900)=420[/tex]