Question

A) If C(x) is the total cost of producing x units of acommodity, then the average cost per unit is c(x)=C(x)/x. Show thatif the average cost is a minimum, then the marginal cost equals theaverage cost.

b) If C(x)=16000+200x+4x^3/2, in dollars, find (i) the cost,average cost, and marginal cost at a production level of 1000units; (ii) the prodution level that will minimize the averagecost; and (iii) the minimum average cost.

Answer 1

a)
`C'(x)=\overline {C}(x)` check below b) The Average Cost \$342.50 Marginal Cost $389.74 ii) The production level that will minimize the average cost: 400 units iii) the minimum average cost $320

Explanation:

a) The Marginal Cost is equal to the Average Cost when its production level is at its minimum. From this question below (letter b) the production is at its minimum at 400 units. For 400 units

`C'(x)=\overline {C}(x)`

Id est:

`\overline {C}(x)=(16000)/(x)+200+4x^{(1)/(2)}\Rightarrow \overline {C}(400)=(16000)/(400)+200+4√(400)\Rightarrow \overline {C}(x)=\$320\\C'(x)=6x^{(1)/(2)}+200\Rightarrow C'(x)=6√(400)+200\Rightarrow C'(x)=\$320`

The Average Cost

`\overline {C}(x)=(C(x))/(x)\Rightarrow \frac{16000+200x+4x^{(3)/(2)}}{x}\\\overline {C}(x)=(16000)/(x)+200+4x^{(1)/(2)}\Rightarrow \\\overline {C}(1000)=(16000)/(1000)+200+4*(1000)^{(1)/(2)}\Rightarrow \overline {C}(1000)=16+200+4(1000)^{(1)/(2)}\Rightarrow \overline {C}(1000)\approx \$342.50`

Marginal Cost

To find the Marginal Cost we need to work with the 1st derivative of the given function, then evaluate for the given production level, 1000 units. The Marginal Cost will return is the cost to produce in this case, approximately the 1001st unit cost.

`C(x)=16000+200x+4x^{(3)/(2)}\\C'(x)=6x^{(1)/(2)}+200`

`C'(x)=6x^{(1)/(2)}+200\Rightarrow C'(x)=6(1000)^{(1)/(2)}+200\Rightarrow C'(x)\approx \$389.74`

ii) The production level that will minimize the average cost is found, when we find out the critical numbers of the function.

The 1st derivative of the Average Cost function:

`\overline {C}(x)=(16000)/(x)+200+4x^{(1)/(2)}\Rightarrow \overline {C}'(x)=(-16000)/(x^(2))+0+\frac{2}{x^{(1)/(2)}}\Rightarrow \overline {C}'(x)=\frac{2x^(2)-16000x^{(1)/(2)}}{x^{(5)/(2)}}`

Setting this equal to zero and solving it to find the production level:

`\overline {C}(x)=(16000)/(x)+200+4x^{(1)/(2)}\Rightarrow \overline {C}'(x)=(-16000)/(x^(2))+0+\frac{2}{x^{(1)/(2)}}\Rightarrow \overline {C}'(x)=\frac{2x^(2)-16000x^{(1)/(2)}}{x^{(5)/(2)}}\Rightarrow \frac{2x^(2)-16000x^{(1)/(2)}}{x^{(5)/(2)}}=0\Rightarrow 2x^(2)-16000x^{(1)/(2)}=0\Rightarrow 2x^(2)-16000√(x)=0\Rightarrow (2x^(2))^2-(16000√(x))^2=0\Rightarrow 4x^(4)-256000000x\Rightarrow x=400`

iii) The minimum average cost

Finally, We'll evaluate average cost function for the minimum production level, to find the minimum average cost.

`\overline {C}(x)=(16000)/(x)+200+4x^{(1)/(2)}\Rightarrow \overline {C}(400)=(16000)/(400)+200+4√(400)\Rightarrow \overline {C}(x)=\$320`

Answer 2

b)

i) Cost: 342491.10 dollars

Average cost: 342.49 dollars per unit

Marginal cost: 389.73 dollars

ii) The average cost is minimized with 400 units

iii) The minimun average cost is 320 dollars per unit

Explanation:

a) If the average cost reaches a minimun when the amount of units is x, then c'(x) = 0. Using the division rule, we can calculate the derivate of c(x) in terms of the derivate of C(x).

`c'(x) = (C'(x)*x - C(x)*1)/(x^2)`

c'(x) = 0 when the numerator is 0, therefore C'(x)*x - C(x) = 0. After a simple calculation, we conclude that
`C'(x) = (C(x))/(x)` , in other words, the marginal cost equals the average cost when the average cost is minimun. That proves (a).

b)
`C(x) = 4x^(3/2) + 200x +16000`

i) The cost, in dollars, is calculated by evaluating C in 1000

`C(1000) = 4*1000^(3/2) + 200*1000 +16000 = 342491.10`

the average cost, in dollars per unit produced, c(1000), is obtained by dividing C(1000) by 1000
`c(1000) = C(1000)/1000 = 342491.10/1000 = 342.49`

In order to calculate the marginal cost, we first derivate C(x):

C'(x) = 4*(3/2) * x^{1/2} + 200 = 6√x + 200

Then, we evaluate the derivate on 1000: C'(1000) = 6√1000 + 200 = 389.73

ii) We need to minimaze c(x) = C(x)/x. Using the item (a), we have to calculate x such that C(x)/x = C'(x), or, equivalently h(x) = C'(x)*x-C(x) = 0.

`h(x) = 6 * x^(3/2)+200x - (4*x^(3/2)+200x+16000) = 2*x^(3/2) -16000`

We obtain that h(x) = 0, only when
`2*x^(3/2) =16000` , thus,
`x^(3/2) =8000` . Therefore
`x = 8000^(2/3) = 400` . We conclude that the production level that will minimize the average cost is 400 units.

iii) The minimun average cost (in dollars per unit) is obtained by evaluating c(x) on 400, which is the value that minimizes the average cost (computed on the previous item). c(400) = C(400)/400 = (16000+200*400+4*400^(3/2))/400 = 40+200+4*√400 = 320.