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) = 2,000 + 160x + 4x3/2 (a) Find the total cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ 288491.11 Correct: Your answer is correct. (b) Find the average cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ 288.49 Correct: Your answer is correct. per unit (c) Find the marginal cost at a production level of 1000 units. (Round your answer to the nearest cent.) $ 349.73 Correct: Your answer is correct. 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:

a) C(1,000)=288,491.11 $

b) c(1,000)=288.49 $/u

c) dC/dx(1,000)=349.74 $/u

d) x=100 u

e) c=220 $/u

Step-by-step explanation:

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

$C(x) = 2,000 + 160x + 4x^{3/2}\\\\C(1,000)=2,000 + 160(1,000) + 4(1,000)^{3/2}\\\\C(1,000)=2,000+160,000+126,491.11= 288,491.11$

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

$c(x)=C(x)/x\\\\c(1,000)=C(1,000)/1,000=288,491.11/1,000=288.49$

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

$\frac{dC}{dx} =0+160+4*(3/2)x^{3/2-1}=160+6x^{1/2}\\\\dC/dx|_{1,000}=160+6*(1,000)^{1/2}=160+189.74=349.74$

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

$c=2,000x^{-1}+160+4x^{1/2}\\\\dc/dx=2,000(-1)x^{-2}+0+4(1/2)x^{-1/2}=0\\\\-2,000x^{-2}+2x^{-1/2}=0\\\\2x^{-1/2}=2,000x^{-2}\\\\x^{-1/2+2}=1,000\\\\x^{3/2}=1,000\\\\x=1,000^{2/3}=100$

(e) What is the minimum average cost?

$c(x) = 2,000/x + 160 + 4x^{1/2}\\\\C(100)=2,000/100 + 160 + 4(100)^{1/2}=20+160+40=220$

Answer 2

Cost functions are used to predict costs based on certain parameters.

• The total cost at a production level of 1000 units is $288491.11
• The average cost at a production level of 1000 units is $288.49
• The marginal cost at a production level of 1000 units is $349.74
• The production level that will minimize the average cost is 100
• The minimum average cost is $220

The cost function is given as:

$\mathbf{C(x) = 2000 + 160x + 4x^\frac{3}{2}}$

(a) The total cost at a production level of 1000 units

This means that, x = 1000

So, we have:

$\mathbf{C(1000) = 2000 + 160 \times 1000 + 4 \times 1000^\frac{3}{2}}$

$\mathbf{C(1000) = 288491.11}$

The total cost at a production level of 1000 units is $288491.11

(b) The average cost at a production level of 1000 units

This is calculated as:

$\mathbf{Average = \frac{Total}{Units}}$

So, we have:

$\mathbf{Average = \frac{C(1000)}{1000}}$

Substitute $\mathbf{C(1000) = 288491.11}$

$\mathbf{Average = \frac{288491.11}{1000}}$

$\mathbf{Average = 288.49111}$

The average cost at a production level of 1000 units is $288.49

(c) The marginal cost at a production level of 1000 units

We start by differentiating the cost function

$\mathbf{C(x) = 2000 + 160x + 4x^\frac{3}{2}}$

So, we have:

$\mathbf{C'(x) = 0 + 160 + \frac 32 \times 4x^{\frac{3}{2} - 1}}$

$\mathbf{C'(x) = 160 + \frac 32 \times 4x^{\frac{1}{2} }}$

$\mathbf{C'(x) = 160 + 6x^{\frac{1}{2} }}$

Substitute, x = 1000

$\mathbf{C'(1000) = 160 + 6 \times 1000^{\frac{1}{2} }}$

$\mathbf{C'(1000) = 349.736 }}$

The marginal cost at a production level of 1000 units is $349.74

(d) The production level of that minimizes the average cost

The average cost is calculated as:

$\mathbf{Average = \frac{Total}{Units}}$

So, we have:

$\mathbf{A(x) = \frac{2000 + 160x + 4x^\frac{3}{2}}{x}}$

Rewrite as:

$\mathbf{A(x) = 2000x^{-1} + 160 + 4x^\frac{1}{2}}$

Differentiate

$\mathbf{A'(x) = -1 \times 2000x^{-2} + 0+ \frac 12 \times 4x^{-\frac{1}{2}}}$

$\mathbf{A'(x) = -2000x^{-2} + 2x^{-\frac{1}{2}}}$

Set to 0

$\mathbf{-2000x^{-2} + 2x^{-\frac{1}{2}} = 0}$

Rewrite as:

$\mathbf{2000x^{-2} = 2x^{-\frac{1}{2}} }$

Divide through by 2

$\mathbf{1000x^{-2} = x^{-\frac{1}{2}} }$

Multiply through by $\mathbf{x^\frac 12}$

$\mathbf{1000x^{-2 + \frac{1}{2}} = x^{-\frac{1}{2} + \frac{1}{2}} }$

$\mathbf{1000x^{-2 + \frac{1}{2}} = x^0 }$

$\mathbf{1000x^{\frac{-4+1}{2}} = 1 }$

$\mathbf{1000x^{\frac{-3}{2}} = 1 }$

Divide both sides by 1000

$\mathbf{x^{\frac{-3}{2}} = \frac{1}{1000} }$

Take multiplicative inverse of both sides

$\mathbf{x^{\frac{3}{2}} = 1000}$

Square both sides

$\mathbf{x^3 = 1000^2}$

Take cube roots of both sides

$\mathbf{x = 1000^\frac 23}$

$\mathbf{x = 100}$

Hence, the production level that will minimize the average cost is 100

Substitute 100 for x in $\mathbf{A(x) = 2000x^{-1} + 160 + 4x^\frac{1}{2}}$, to get the minimum average cost

$\mathbf{A(100) = 2000 \times 100^{-1} + 160 + 4 \times 100^\frac{1}{2}}$

$\mathbf{A(100) = 20 + 160 + 4 \times 10}$

$\mathbf{A(100) = 20 + 160 + 40}$

$\mathbf{A(100) = 220}$

The minimum average cost is $220

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