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# Calculus

Applications of the Derivative

# Optimization Problems in Economics

Page 1
Problems 1-4
Page 2
Problems 5-8

The examples below describe various situations that can occur in business and economic activities.
In each of these problems it is necessary to construct a function that establishes a relationship between two economic variables (for example, between the volume of production and revenue) and investigate it for extreme values. Optimization problems of this type can be effectively solved using derivatives.

## Solved Problems

Click on problem description to see solution.

### ✓Example 7

Suppose that the production output $$Q$$ depends on the number of workers $$L$$ and this dependence is described by a function $$Q\left( L \right)$$ (Figure $$1$$). Show that if the derivatives satisfy the conditions

$Q’\left( L \right) > 0,\;\;\;Q”\left( L \right) < 0,$

then there is an optimal number of workers $${L^*},$$ when the profit is maximized.

### ✓Example 8

Two cities $$A$$ and $$B$$ are located at the distance of $$a\,\text{miles}$$ from each other and are connected by a straight railroad. For the transportation of goods from city $$A$$ to city $$C$$ located $$b\,\text{miles}$$ from the railroad it is necessary to build a highway adjacent to the railway line (Figure $$2$$). Find the point $$S$$ on the railroad where the highway should be built to provide the most cost-effective transportation of goods. The shipping cost is $$p\,\$$ per ton-mile by rail and $$q\,\$$ per ton-mile by truck.

### Example 1.

A plant produces and sells semiconductor devices. The cost per one unit (also known as the unit cost) depends on the volume of production and includes a fixed part $$1000$$ ($/device) and a variable part $$2n$$ ($/device), where $$n$$ is the number of units produced per month. The price of the device, in turn, depends on the volume of production according to the law $$p\left( n \right) = 10000 – n$$ (\$/device). Determine at what volume of production the profit will be highest?

#### Solution.

The income from the sale of units manufactured during a month is

$R\left( n \right) = np\left( n \right) = n\left( {10000 – n} \right).$

The monthly expenses are given by

$C\left( n \right) = n\left( {1000 + 2n} \right).$

Then the profit is determined by the formula

${P\left( n \right) = R\left( n \right) – C\left( n \right) } = {n\left( {10000 – n} \right) – n\left( {1000 + 2n} \right) } = {10000n – {n^2} – 1000n – 2{n^2} } = {9000n – 3{n^2}.}$

Investigate extreme values of the profit function. Assuming that $$n$$ is a real number and differentiating with respect to $$n,$$ we get:

${P’\left( n \right) = {\left( {9000n – 3{n^2}} \right)^\prime } = 9000 – 6n = 0,\;\;}\Rightarrow { n = \frac{{9000}}{6} = 1500.}$

Calculate also the second derivative:

$P^{\prime\prime}\left( n \right) = {\left( {9000 – 6n} \right)^\prime } = – 6 \lt 0.$

Since the second derivative is negative everywhere, the solution $$n = 1500$$ is a maximum point. Thus, production of $$1500$$ devices per month provides the highest profit for the company.

### Example 2.

To produce $$x$$ units of some product a company spends

$C\left( x \right) = a{x^2} + bx\;\left(\\right),$

where $$a$$ and $$b$$ are real numbers. The product is sold at price $$p\,\$$ per unit. Determine the sales volume at which profit reaches its maximum.

#### Solution.

When selling $$x$$ units of the product, the company has income equal to

$R\left( x \right) = px.$

Hence, the profit of the company is

${P\left( x \right) = R\left( x \right) – C\left( x \right) } = {px – \left( {a{x^2} + bx} \right) } = {\left( {p – b} \right)x – a{x^2}.}$

Find the derivative of the function $$P\left( x \right):$$

${P’\left( x \right) } = {{\left[ {\left( {p – b} \right)x – a{x^2}} \right]^\prime } } = {p – b – 2ax.}$

The derivative is zero at the point

${P’\left( x \right) = 0,\;\;}\Rightarrow {p – b – 2ax = 0,\;\;}\Rightarrow {2ax = p – b,\;\;}\Rightarrow {x = \frac{{p – b}}{{2a}}.}$

Consider the second derivative:

$P^{\prime\prime}\left( x \right) = {\left( {p – b – 2ax} \right)^\prime } = – 2a \lt 0.$

Since the second derivative is negative, then the point $$x = {\large\frac{{p – b}}{{2a}}\normalsize}$$ is the maximum point, i.e. the company will have the maximum profit at the given sales volume.

### Example 3.

A company sells its products at unit price of $$100\,\$$ if the lot size does not exceed $$5000$$ units. At higher volume the company gives a discount of $$5\,\$$ for each additional thousand exceeding the level of $$5000.$$ Determine the order volume at which the company has the largest income.

#### Solution.

Let $$x$$ be the number of products in a lot. If $$x \le 5000,$$ the unit price is $$100\,\$$ by condition. If $$x \gt 5000,$$ the price is calculated by the formula

${p\left( x \right) } = {100 – 5 \cdot \frac{{x – 5000}}{{1000}} } = {100 – 0,005x + 25 } = {125 – 0,005x.}$

In the first case, for $$x \le 5000,$$ the maximum income is reached at $$x = 5000.$$ It is equal to
${R_1} = 5000 \cdot 100 = 500000\;\left(\\right).$
In the second case, for $$x \gt 5000,$$ the income is defined by the following function:

${{R_2} = R\left( x \right) } = {xp\left( x \right) } = {x\left( {125 – 0,005x} \right) } = {125x – 0,005{x^2}\;\left(\\right).}$

Find the derivative:

${R’\left( x \right) } = {{\left( {125x – 0,005{x^2}} \right)^\prime } } = {125 – 0,01x.}$

By equating it to zero, we determine the critical point:

${R’\left( x \right) = 0,\;\;}\Rightarrow { 125 – 0,01x = 0,\;\;}\Rightarrow {x = \frac{{125}}{{0,01}} = 12500.}$

Note that the second derivative of the function $$R\left( x \right)$$ is always negative:

${R^{\prime\prime}\left( x \right) } = {{\left( {125 – 0,01x} \right)^\prime } } = { – 0,01 \lt 0.}$

So the critical point we found corresponds to the maximum of the function $$R\left( x \right).$$ Thus, the company’s income is maximized when selling lots of $$x = 12500$$ units. The maximum income will be equal to

${{R_{\max }} } = {125 \cdot 12500 – 0,005 \cdot {12500^2} } = {781250\;\left(\\right).}$

### Example 4.

A certain country uses a progressive tax system. The amount of tax consists of a linear part proportional to the income and a nonlinear part depending on the income by a power law. The total amount of tax is determined by the formula

$T\left( W \right) = aW + {\left( {bW + c} \right)^p},$

where $$W$$ is the income; $$p$$ is the exponent, $$a, b, c$$ are some positive numbers. At what level of income the tax rate will be minimal?

#### Solution.

The tax rate $$r$$ is calculated by the formula

${r\left( W \right) = \frac{{T\left( W \right)}}{W} } = {\frac{{aW + {{\left( {bW + c} \right)}^p}}}{W} } = {a + \frac{{{{\left( {bW + c} \right)}^p}}}{W}.}$

Investigate it for extreme values. Find the derivative:

${r’\left( W \right) = {\left[ {a + \frac{{{{\left( {bW + c} \right)}^p}}}{W}} \right]^\prime } } = {\frac{{p{{\left( {bW + c} \right)}^{p – 1}} \cdot W – {{\left( {bW + c} \right)}^p} \cdot 1}}{{{W^2}}} } = {\frac{{{{\left( {bW + c} \right)}^{p – 1}}\left[ {pW – \left( {bW + c} \right)} \right]}}{{{W^2}}} } = {\frac{{{{\left( {bW + c} \right)}^{p – 1}}\left[ {\left( {p – b} \right)W – c} \right]}}{{{W^2}}}.}$

It can be seen that function $$r\left( W \right)$$ has three critical points, but since the coefficients $$b, c > 0,$$ then a meaningful solution exists only in the following point:

${\left( {p – b} \right)W – c = 0,\;\;}\Rightarrow {W = \frac{c}{{p – b}}.}$

When passing through this value the derivative changes sign from minus to plus. Consequently, the function $$r\left( W \right)$$ reaches a minimum here, i.e. the tax rate will be the least at this income level.

Page 1
Problems 1-4
Page 2
Problems 5-8