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.

Question

C(x) = 81,000 + 170x + 6x^3/2

(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.)

(c) Find the marginal cost at a production level of 1000 units. (Round your answer to the nearest cent.)

(d) Find the production level that will minimize the average cost. (Round your answer to the nearest whole number.)

(e) What is the minimum average cost? (Round your answer to the nearest dollar.)

Answer 1

Simon

1. Here just substitute x=1000 into C(x) giving $ 440736.6596, or $ 440736.66 to nearest cent
2. Find C(1000) / 1000. Using C(1000) from (1) gives 440.7366596, or $ 440.74 to nearest cent
3. Marginal cost means the cost of producing one unit more, given that 1000 have been produced. There are a couple of ways of finding this : C(1001) - C(1000) or dC/dx when x=1000. As this is calculus let's use the second way.

dC/dx = 170 + 6. 3/2 x^1/2 (using d/dx xⁿ= n x^n-1 )

= 170 + 9 x ^1/2. When x=1000 this evaluates to $ 454.60 to the nearest cent.

(d). Average cost is C(x)/x = 81000/x + 170 + 6x^1/2

Let's call this function A(x).

To minimise this we set dA/dx = 0.

dA/dx = -81000/x²+3x^-1/2=-81000/x²+3/√x

For this to be zero, 3/√x = 81000 / x², giving 27000 = x ^3/2. log (x) = log (27000) / (3/2) so

x=10^{log (27000)/(3/2)} = 900.

So the production level minimising the average cost will be x = 900

As for the last part you need to find C(900)/900 here, which is $ 440

Mike