# Optimization Problems in Economics

In business and economics there are many applied problems that require optimization. For example, in any manufacturing business it is usually possible to express profit as function of the number of units sold. Finding a maximum for this function represents a straightforward way of maximizing profits.

The problems of such kind can be solved using differential calculus.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A game console manufacturer determines that in order to sell $$x$$ units, the price per one unit (in dollars) must decrease by the linear law (the demand function) ${p\left( x \right) = 500 - 0.1x \;\left( {\frac{\ }{\text{device}}} \right)}.$ The manufacturer also determines that the cost depends on the volume of production and includes a fixed part $$100,000 \left( {\} \right)$$ and a variable part $$100x$$, that is $C\left( x \right) = {100000} + {100x}.$ What price per unit must be charged to get the maximum profit?

### Example 2

The demand function for a certain commodity is $p\left( x \right) = 10 - 0.001x,$ where $$p$$ is measured in dollars and $$x$$ is the number of units produced and sold. The total cost of producing $$x$$ items is $C\left( x \right) = 50 + 5x.$ Determine the level of production that maximizes the profit.

### Example 3

The demand function for a certain product is linear and defined by the equation $p\left( x \right) = 10 - \frac{x}{2},$ where $$x$$ is the total output. Find the level of production at which the company has the maximum revenue.

### Example 4

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 $p\left( n \right) = 10000 - n\; \left(\/device\right).$ Determine at what volume of production the profit will be highest?

### Example 5

A shop sells pies for $$\5$$ each. The daily cost function has the form $C\left( x \right) = x + 10 + 0.01{x^2},$ where $$x$$ is the number of pies sold on a typical day. Find the value of $$x$$ that will maximize the daily profit.

### Example 6

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.

### Example 7

The revenue for a certain product is given by the equation $R\left( x \right) = 100 - \frac{{400}}{{x + 5}} - x,$ where $$x$$ is the number of produced items. Find the value of $$x$$ that results in maximum revenue.

### Example 8

The production cost per a period of time is given by the quadratic function $C\left( x \right) = a + b{x^2},$ where $$a,b$$ are some positive real numbers and $$x$$ represents the number of units. Find the minimal average cost. (The average cost is the total cost divided by the number of units produced.)

### Example 1.

A game console manufacturer determines that in order to sell $$x$$ units, the price per one unit (in dollars) must decrease by the linear law (the demand function) ${p\left( x \right) = 500 - 0.1x \;\left( {\frac{\ }{\text{device}}} \right)}.$ The manufacturer also determines that the cost depends on the volume of production and includes a fixed part $$100,000 \left( {\} \right)$$ and a variable part $$100x$$, that is $C\left( x \right) = {100000} + {100x}.$ What price per unit must be charged to get the maximum profit?

Solution.

The total revenue is given by

$R\left( x \right) = xp\left( x \right) = x\left( {500 - 0.1x} \right) = 500x - 0.1{x^2}$

The profit is determined by the formula

$P\left( x \right) = R\left( x \right) - C\left( x \right) = 500x - 0.1{x^2} - \left( {100000 + 100x} \right) = 400x - 0.1{x^2} - 100000$

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

$P^\prime\left( x \right) = \left( {400x - 0.1{x^2} - 100000} \right)^\prime = 400 - 0.2x$

There is one critical value:

$P^\prime\left( x \right) = 0,\;\; \Rightarrow 400 - 0.2x = 0,\;\; \Rightarrow x = 2000.$

We use the Second Derivative Test to classify the critical point.

$P^{\prime\prime}\left( x \right) = \left( {400 - 0.2x} \right)^\prime = - 0.2 \lt 0$

Since $$P^{\prime\prime}\left( x \right)$$ is negative, $$x = 2000$$ is a point of maximum.

Hence, the profit is maximized when $$2000$$ game consoles are sold.

In this case, the price per unit is equal to

$p\left( {x = 2000} \right) = 500 - 0.1 \cdot 2000 = 300\,\left( {\frac{\ }{\text{device}}} \right).$

### Example 2.

The demand function for a certain commodity is $p\left( x \right) = 10 - 0.001x,$ where $$p$$ is measured in dollars and $$x$$ is the number of units produced and sold. The total cost of producing $$x$$ items is $C\left( x \right) = 50 + 5x.$ Determine the level of production that maximizes the profit.

Solution.

The profit function is given by

$P\left( x \right) = xp\left( x \right) - C\left( x \right) = x\left( {10 - 0.001x} \right) - \left( {50 + 5x} \right) = 10x - 0.001{x^2} - 50 - 5x = 5x - 0.001{x^2} - 50$

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

$P^\prime\left( x \right) = \left( {5x - 0.001{x^2} - 50} \right)^\prime = 5 - 0.002x,$

so the critical point is

$x = \frac{5}{{0.002}} = 2500.$

Since the second derivative of $$P\left( x \right)$$ is negative, $$x = 2500$$ is a point of maximum.

Hence, the company has the largest profit when $$x = 2500.$$

### Example 3.

The demand function for a certain product is linear and defined by the equation $p\left( x \right) = 10 - \frac{x}{2},$ where $$x$$ is the total output. Find the level of production at which the company has the maximum revenue.

Solution.

The revenue is defined by the formula

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

Hence,

$R\left( x \right) = x\left( {10 - \frac{x}{2}} \right) = 10x - \frac{{{x^2}}}{2}.$

We see that $$R\left( x \right)$$ is a parabola curved downward. It has a maximum at the following point:

$R^\prime\left( x \right) = \left( {10x - \frac{{{x^2}}}{2}} \right)^\prime = 10 - x;$
$R^\prime\left( x \right) = 0,\;\; \Rightarrow 10 - x = 0,\;\; \Rightarrow x = 10.$

As the second derivative of the function $$R\left( x \right)$$ is negative, the point $$x = 10$$ is a point of maximum.

Thus, the maximum revenue is attained at the production rate $$x = 10.$$

### Example 4.

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 $p\left( n \right) = 10000 - n\; \left(\/device\right).$ 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 5.

A shop sells pies for $$\5$$ each. The daily cost function has the form $C\left( x \right) = x + 10 + 0.01{x^2},$ where $$x$$ is the number of pies sold on a typical day. Find the value of $$x$$ that will maximize the daily profit.

Solution.

The profit function is written as

$P\left( x \right) = R\left( x \right) - C\left( x \right),$

where the revenue $$R\left( x \right)$$ is given by $$R\left( x \right) = xp$$ ($$p$$ is the price per one pie). Then

$P\left( x \right) = xp - C\left( x \right) = 5x - \left( {x + 10 + 0.01{x^2}} \right) = 4x - 0.01{x^2} - 10$

The derivative of $$P\left( x \right)$$ is

$P^\prime\left( x \right) = \left( {4x - 0.01{x^2} - 10} \right)^\prime = 4 - 0.02x;$

Determine the point at which the derivative is zero:

$P^\prime\left( x \right) = 0,\;\; \Rightarrow 4 - 0.02x = 0,\;\; \Rightarrow x = \frac{4}{{0.02}} = 200$

Notice that the second derivative is negative:

$P^{\prime\prime}\left( x \right) = \left( {4 - 0.02x} \right)^\prime = - 0.02 \lt 0$

herefore, $$x = 200$$ is a point of maximum, so the largest profit is attained at $$x = 200.$$

### Example 6.

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 $$x = {\frac{{p - b}}{{2a}}}$$ is the maximum point, i.e. the company will have the maximum profit at the given sales volume.

### Example 7.

The revenue for a certain product is given by the equation $R\left( x \right) = 100 - \frac{{400}}{{x + 5}} - x,$ where $$x$$ is the number of produced items. Find the value of $$x$$ that results in maximum revenue.

Solution.

We differentiate the function $$R\left( x \right)$$ to determine its critical point.

$R^\prime\left( x \right) = \left( {100 - \frac{{400}}{{x + 5}} - x} \right)^\prime = \frac{{400}}{{{{\left( {x + 5} \right)}^2}}} - 1;$
$R^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{400}}{{{{\left( {x + 5} \right)}^2}}} - 1 = 0,\;\; \Rightarrow 400 = {\left( {x + 5} \right)^2},\;\; \Rightarrow x + 5 = 20,\;\; \Rightarrow x = 15.$

Using the First Derivative Test, we can verify that $$x = 15$$ is a point of maximum.

Hence, the maximum revenue occurs when $$x = 15.$$

### Example 8.

The production cost per a period of time is given by the quadratic function $C\left( x \right) = a + b{x^2},$ where $$a,b$$ are some positive real numbers and $$x$$ represents the number of units. Find the minimal average cost. (The average cost is the total cost divided by the number of units produced.)

Solution.

By definition, the average cost $${\overline C}$$ is written in the form

$\overline C\left( x \right) = \frac{{C\left( x \right)}}{x} = \frac{{a + b{x^2}}}{x} = \frac{a}{x} + bx.$

Take the derivative and set it equal to zero to find the critical points:

$\overline C^\prime\left( x \right) = \left( {\frac{a}{x} + bx} \right)^\prime = - \frac{a}{{{x^2}}} + b = \frac{{b{x^2} - a}}{{{x^2}}};$
$\overline C^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{b{x^2} - a}}{{{x^2}}} = 0,\;\; \Rightarrow b{x^2} - a = 0,\;\; \Rightarrow x = \sqrt {\frac{a}{b}} .$

By the First Derivative Test, we identify that this point is a point of minimum.

Calculate the minimal average cost:

${{\overline C}_{\min }} = \overline C\left( {\sqrt {\frac{a}{b}} } \right) = \frac{a}{{\sqrt {\frac{a}{b}} }} + b\sqrt {\frac{a}{b}} = \sqrt {ab} + \sqrt {ab} = 2\sqrt {ab} .$

See more problems on Page 2.