在经济任务中使用函数导数

Maka Lomtadze
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摘要

本文着重讨论了数学方法在经济学中的应用,特别是讨论了容易用衍生工具解决的经济问题。这篇文章的目的是向学生展示使用数学方法解决经济问题的方法和机会。为此,本文对几个经济任务进行了详细的讨论和分析,既有趣又易于学生掌握。我把函数的导数看作是变化率,并引入了这样的定义:函数f在某一点关于x的瞬时变化率,如果它存在,就称为导数。在这个定义的帮助下,我已经讨论和解释了任务1:假设某种产品在一段时间内的产量增加用一个函数来描述,人口增长用以下函数来描述:其中为初始时期的年数,那么这些产品的人均产量用函数给出:求产品生产的增长率。通过解决这个任务,我得出结论,一年后人均产品产量增加。在接下来的任务中,我使用了用导数求函数极值的方法,即将一阶导数等于0,找到临界点,然后借助二阶导数确定函数的极值。我讨论了任务2:对于生产X量的产品,公司计划一个由公式计算的成本。什么产量的平均成本最小?求出这一小笔费用的数值。通过求解这个问题,我得出了当产量为一个单位时,平均成本的给定函数取最小值,这个值等于:,即生产该产量时的边际成本。我讨论了任务3:如果衍生成本函数已知,为了使公司利润最大化,应该销售多少产品?以及回报函数:在这里我得出结论:如果销售了600件产品,那么公司的利润将是最大的,并且它将在数字上相等。在文章的最后,我讨论了这样一个任务4的应用优化。制造一个2升的圆柱形罐子最少需要多少材料?我得出的结论是:如果我们取底部半径的2倍的高度,我们将花费最少的材料来得到一个圆柱形容器。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
USING A FUNCTION DERIVATIVE IN ECONOMIC TASKS
The article focuses on the application of mathematical methods in economics, in particular discussing economic problems that are easily solved using derivatives. The purpose of the article is to show students the way and opportunity to use mathematical methods to solve economic problems. To this end, the article discusses and analyzes several economic tasks in detail, which will be interesting and easy for students to master. I considered the derivative of a function as the rate of change and introduced the definition: The instantaneous rate of change of the function f with respect to x at a point is called the derivative if it exists. With the help of this definition I have discussed and explained Task 1: Suppose that the increase in production of a certain product over a period of time is described by a function And population growth is described by the following function: Where is the number of years from the initial period, then the production of these products per capita is given by the function: Find the growth rate of product production. By solving this task I came to the conclusion that after a year the production of products per capita increases. In the following tasks I used the method of finding the extremum values of a function using a derivative, namely I equated the first-order derivative to 0, found the critical points, and with the help of the second-order derivative I determined the extremes of the function. I discussed task 2: For the production of X volume of products, the firm plans a cost that is calculated by the formulan . For what volume of production will the average cost be the smallest? Find the numerical value of this small expense. Solving this problem, I came to the conclusion that the given function of the average cost takes on the least value when the output volume is a unit , and this value is equal to: which is the marginal cost when producing the volume output. I discussed Task 3: How many products should be sold in order for a firm to profit maximally if the derivative cost function is known: And return function: Here I came to the conclusion: if 600 units of the product are sold, then the firm's profit will be maximum and it will be numerically equal At the end of the article I discussed such a task 4 of applied optimization. What is the minimum amount of material needed to make a 2 liter cylindrical jar? Where I came to the conclusion: the smallest amount of material will be spent to get a cylindrical vessel if we take the height 2 times the radius of the base.
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