Free Raw Materials And Basic Materials - $8000 Research Paper Example

Writing Assignment

1)
I will provide a fictional example. Assume I want to produce electric kettles, the price of such devices is usually between $15 and $300. Assume that the cost structure of calculation per 1000 products is as follows:

Fuel and electricity for technological purposes - $1000.
Wages of production workers - $5000.
Accruals for wages - 40% of the wages of production workers
Production costs - 10% of the wages of production workers.
General Expenses - 20% of the wages of production workers.
Transportation costs and packaging - 5% to production costs.
Calculate in absolute terms, indirect costs, and the data as a percentage of wages of production workers on 1000 units:
Accruals for salaries = $5000 * 40%: 100% =$2000;
Overhead expenses = $5000 * 10%: 100% = $500;
General expenses = $5000 * 20%: 100% = $1000.
2. Determine the production cost as the sum of the costs of articles 1-6.
Cost of production per 1000 units = $8000+$1000+$5000+$2000+$500+$1000=$17500
3. The costs of transport and packaging = $17500 · 5%: 100% = $875.
4. The total cost of 1000 units = $17500 + $875 = $18375; total cost of one product = $18.38.
2)

If fixed cost is $7,500 per month, then the cost function is:

Cx=18.38x+7500

Where x represents the amount of produced items.

If I can afford $150,000 cost at max, then:
18.38x+7500≤150000x≤150000-750018.38=7752.99

Hence, we can produce not more than 7752 kettles.

Now, the demand function is:
x=9600-30pp=9600-x30

Then the revenue function is:

Rx=x*p=9600x-x230

Find the derivative and maximum and minimum points:

R'x=320-x15
The maximum is at x=4800, and it is equal to $768,000. The minimum value is at 0.

Now the profit function is:

Px=Rx-Cx=9600x-x230-18.38x+7500=-x230+301.62x+7500
-x230+301.62x+7500>0
I use www.wolframaplha.com to solve the equation:
-x230+301.62x+7500>0

The solution is: 0

Hence, we have to produce not more than 9073.4 or the cost will be higher than the revenue.
3)
MC=TC'=18.38-cost per unit18380 for 1000 units

The profit for 1000 units is:

-1000230+301.62*1000+7500=$275,786.67
-x230+301.62x+7500→max
I use Wolfram Alpha once again and get approximate solution:
Maximum is at x=4524.3 and it is equal to $689,810.
Hence, if we produce 1000 units, we have to increase the production to 4524.3 units.

Summary

In this paper we have demonstrated the basics of economics of production. We demonstrated such concepts as cost per unit, marginal cost, revenue, profit and demand functions and others. We have chosen an item to produce (kettle) and perform all necessary operations to find the optimal volume of production. As a result, the best way for our company is to produce 4524 kettles per month to satisfy the demand and obtain the maximum profit.
All the work we have done shows that mathematics is very useful and significant to solve real world problems, especially in economics.

Works Cited

Frisch, R., Theory of Production, Drodrecht: D. Reidel, 1965.
Ferguson, C. E., The Neoclassical Theory of Production and Distribution, London: Cambridge Univ. Press, 1969.
Pindyck, R., and Rubinfeld, D., Microeconomics, 5th ed., Prentice-Hall, 2001.
Nicholson: Microeconomic Theory 9th ed. Page 238 Thomson 2005