Question

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 + 130x + 4x3/2 (a) Find the total cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ (b) Find the average cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ per unit (c) Find the marginal cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ per unit (d) Find the production level that will minimize the average cost. (Round your answer to the nearest whole number.) units (e) What is the minimum average cost? (Round your answer to the nearest dollar.) $ per unit

Answer 1

Answer:

Step-by-step explanation:

Given that:

If C(x) =  the cost of producing x units of a commodity

Then;

then the average cost per unit is c(x)  = $\dfrac{C(x)}{x}$

We are to consider a given function:

$C(x) = 54,000 + 130x + 4x^{3/2}$

And the objectives are to determine the following:

a) the total cost at a production level of 1000 units.

So;

If C(1000) = the cost of producing 1000 units of a commodity

$C(1000) = 54,000 + 130(1000) + 4(1000)^{3/2}$

$C(1000) = 54,000 + 130000 + 4( \sqrt[2]{1000^3} )$

$C(1000) = 54,000 + 130000 + 4(31622.7766)$

$C(1000) = 54,000 + 130000 + 126491.1064$

$C(1000) = 310491.1064$

$\mathbf{C(1000) \approx 310491.11 }$

(b) Find the average cost at a production level of 1000 units.

Recall that :

the average cost per unit is c(x)  = $\dfrac{C(x)}{x}$

SO;

$c(x) =\dfrac{(54,000 + 130x + 4x^{3/2})}{x}$

Using the law of indices

$c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$

$c(1000) = \dfrac{54000}{1000}+ 130 + {4(1000)^{1/2}}$

c(1000) =$ 310.49 per unit

(c) Find the marginal cost at a production level of 1000 units.

The marginal cost  is C'(x)

Differentiating  C(x) = 54,000 + 130x + 4x^{3/2} to get  C'(x) ; we Have:

$C'(x) = 0 + 130 + 4 \times \dfrac{3}{2} \ x^{\dfrac{3}{2}-1}$

$C'(x) = 0 + 130 + 2 \times \ {3} \ x^{\frac{1}{2}}$

$C'(x) = 0 + 130 + \ {6}\ x^{\frac{1}{2}}$

$C'(1000) = 0 + 130 + \ {6} \ (1000)^{\frac{1}{2}}$

$C'(1000) = 319.7366596$

$\mathbf{C'(1000) = \ 319.74 \ per \ unit}$

(d)  Find the production level that will minimize the average cost.

the average cost per unit is c(x)  = $\dfrac{C(x)}{x}$

$c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$

the production level that will minimize the average cost is c'(x)

differentiating $c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$ to get c'(x); we have

$c'(x)= \dfrac{54000}{x^2} + 0+ \dfrac{4}{2 \sqrt{x} }$

$c'(x)= \dfrac{54000}{x^2} + 0+ \dfrac{2}{ \sqrt{x} }$

Also

$c''(x)= \dfrac{108000}{x^3} -x^{-3/2}$

$c'(x)= \dfrac{54000}{x^2} + \dfrac{4}{2 \sqrt{x} } = 0$

$x^2 = 27000\sqrt{x}$

$\sqrt{x} (x^{3/2} - 27000) =0$

x= 0;  or  $x= (27000)^{2/3}$ = $\sqrt[3]{27000^2}$ = 30² = 900

Since  production cost can never be zero; then the production cost = 900 units

(e) What is the minimum average cost?

the minimum average cost of c(900) is

$c(900) =\dfrac{54000}{900} + 130 + 4(900)^{1/2}$

c(900) = 60 + 130 + 4(30)

c(900) = 60 +130 + 120

c(900) = $310 per unit