Economic Models

Section 2.7 Economic Models

Subsection 2.7.1 Demand and Supply Functions

In a market economy that has few or no restrictions and regulations on buyers and sellers, the consumer demand for a particular commodity is dependent on the commodity's unit price. The relationship between a unit price and the quantity demanded is articulated by a so-called demand equation and its graph is referred to as a demand curve. In general, the quantity demanded of a commodity increases as the commodity's unit price decreases, and vice versa.

Definition 2.40. Demand Function.

A demand function is defined by $p = f(x)\text{,}$ where $p$ measures the unit price and $x$ measures the number of units of the commodity in question, and is generally characterized as a decreasing function of x; that is, $p = f(x)$ decreases as $x$ increases. Since both $x$ and $p$ assume only nonnegative values, the demand curve is that part of the graph of $f(x)$ that lies in the first quadrant (Figure 2.7).

Figure 2.7. Example of a supply curve (in blue) and a demand curve (in red). The point of intersection $(x_0, p_0)$ corresponds to market equilibrium.

In a market in which a large number of producers compete with each other to satisfy the wants and needs of a large number of consumers, the unit price of a commodity is dependent on the commodity's availability in the market. The relationship between a unit price and the quantity supplied is articulated by a so-called supply equation and its graph is referred to as a supply curve. In general, an increase or decrease in the commodity's unit price induces the producer to respectively increase or decrease the supply of the commodity.

Definition 2.41. Supply Function.

A supply function defined by $p = f(x)$ with $p$ and $x$ as before is generally characterized as an increasing function of $x\text{;}$ that is, $p = f(x)$ increases as $x$ increases. Since both $x$ and $p$ assume only nonnegative values, the supply curve is that part of the graph of $f(x)$ that lies in the first quadrant (Figure 2.7)

In a competitive market, the price of a commodity will eventually settle at a certain level, namely when the supply of the commodity will be equal to the demand for it. Simply put, if the price is too high, then the consumer will not buy as much of the product as is being supplied; conversely, if the price is too low, then the supplier will not produce as much of the product as is being demanded. When the quantity produced is equal to the quantity demanded, then a so-called market equilibrium is achieved. The number of units at which demand equals supply is the equilibrium quantity, the corresponding price per unit is the equilibrium price, and the corresponding ordered pair gives the equilibrium point. f

Definition 2.42. Equilibrium Point.

The equilibrium point $(x,p)$ is defined by

\begin{equation*} (x, p) = (\text{ equilibrium quantity } , \text{ equilibrium price } ) \end{equation*}

and provides the quantity $x$ and price $p$ at which the demand equals the supply.

Since the supply of a product is equal to the demand for it under market equilibrium, this market equilibrium must be the point at which the demand curve and the supply curve intersect. In Figure 2.7, the point $(x_0,p_0)$ represents (equilibrium quantity, equilibrium price). Since the point $(x_0, p_0)$ lies on the supply curve as well as the demand curve, it satisfies both the supply equation and the demand equation. Thus, to find the market equilibrium, we solve the demand and supply equations simultaneously for $x$ and $p\text{,}$ which will yield the point $(x_0,p_0)\text{.}$ Mathematically speaking one or both of the values may be negative when solving the system of two equations; however, for the solutions to be meaningful, both $x_0$ and $p_0$ must be positive.

Example 2.43. Supply-Demand.

The demand function for a certain commercial product is given by

\begin{equation*} p = d(x) = -0.01x^{2}-0.2x+8 \end{equation*}

and the corresponding supply function is given by

\begin{equation*} p = s(x) = 0.01x^{2} + 0.1x + 3 \end{equation*}

where $p$ is expressed in dollars and $x$ is measured in units of a hundred. Find the market equilibrium.

Solution

We solve the following system of equations:

\begin{equation*} \begin{split} p \amp = -0.01x^{2} - 0.2x + 8\\ p \amp = 0.01x^{2} + 0.1x + 3 \end{split} \end{equation*}

Subtracting the first equation from the second equation, we obtain

\begin{equation*} -0.01x^{2} - 0.2x + 8 = 0.01x^{2} + 0.1x + 3 \end{equation*}

which is equivalent to

\begin{equation*} \begin{split} 0.02x^{2} + 0.3x - 5 \amp = 0\\ 2x^{2} + 30x - 500 \amp = 0 \\ x^{2} + 15x - 250 \amp = 0 \\ (x+25)(x-10) \amp = 0 \end{split} \end{equation*}

Thus, $x = -25$ or $x = 10\text{.}$ Since the number of units $x$ must be nonnegative, the root $x = -25$ is rejected. Therefore, the equilibrium quantity is $1000$ units. The equilibrium price is given by

\begin{equation*} p = 0.01(10^{2}) + 0.1(10) + 3 = 5 \end{equation*}

or $5 per unit. The graphs of the demand and supply functions and their intersection point are shown below.

Subsection 2.7.2 Cost, Revenue and Profit Functions

When a problem arising from a practical situation is being modelled mathematically, then this often leads to an expression that involves the combination of functions. For example, modelling the costs incurred in running a business. Here, we differentiate between costs that remain more or less constant regardless of the firm's level of activity, namely the fixed costs or constant costs, and costs that vary with production or sales, namely the variable costs or marginal costs. Rental fees and executive salaries are examples of fixed costs, while wages and purchases of raw materials are examples of variable costs. Therefore, in order to consider the total cost of operating a business, typically denoted by $C\text{,}$ we must sum the variable costs and the fixed costs.

Definition 2.44. Linear Cost Function.

The linear cost function $C$ is given by

\begin{equation*} C(x) = mx + b \end{equation*}

where $m$ is the variable cost per unit, $b$ is the overall fixed cost, and $x$ is the number of units produced.

The average cost of manufacturing $x$ units of a certain product is obtained by dividing the total production cost by the number of units manufactured.

Definition 2.45. Average Cost Function.

Suppose $C(x)$ is the total cost function. Then the average cost function, denoted by $\overline{C}(x)$ — read “$C$ bar of $x$” — is given by

\begin{equation*} \overline{C}(x) = \dfrac{C(x)}{x}\text{.} \end{equation*}

The revenue realized by a company from the sale of $x$ units of a certain commodity is given by the so-called revenue function, typically denoted by $R\text{.}$ Then a simple model is revenue = (price per unit)$\cdot$(number of units produced or sold). However, the market in which the company operates dictates the price that the company can demand for the product. Recall that in a competitive market environment the price is determined by market equilibrium. In other words, if the company is one of many, then none of them is able to dictate the price of the commodity. But in a monopolistic market, the company is the sole supplier of the product and can therefore manipulate the price of the commodity by controlling the supply. The unit selling price $p$ of the commodity and the quantity $x$ of the commodity demanded are related to each other by the demand function. This leads to the following definition of the revenue function.

Definition 2.46. Revenue Function.

The Revenue function $R$ is given by

\begin{equation*} R(x) = px = xf(x) \end{equation*}

where $p$ is the unit selling price of the commodity, $x$ is the quantity of the commodity demanded, and $f$ is the demand function.

The profit realized by a company in operating a business is the difference between the total revenue realized and the total cost incurred.

Definition 2.47. Profit Function.

The Profit function $P$ is given by

\begin{equation*} P(x) = R(x)-C(x) \end{equation*}

where $R$ is the revenue function, $C$ is the cost function, and $x$ is the quantity of the commodity sold.

Example 2.48. Cost Functions.

Suppose a certain company has a monthly fixed cost of $$10,000$ and a variable cost of

\begin{equation*} -0.0001x^{2} + 12x \ \ \ 0 \leq x \leq 40,000 \end{equation*}

dollars, where $x$ denotes the number of units manufactured per month. Find a function $C$ that gives the total cost incurred by the company in the manufacture of $x$ units.

Solution

The company's monthly fixed cost is $$10,000\text{,}$ regardless of the level of production. The total cost will be the sum of the variable and fixed costs,

\begin{equation*} C(x) = -0.0001x^{2} + 12x + 10000 \ \ \ 0\leq x \leq 40,000 \end{equation*}

Example 2.49. Profit Functions.

Suppose the total revenue realized by the same company from Example 2.48 from the sale of $x$ units is given by the total revenue function

\begin{equation*} R(x) = -0.0005x^{2} + 22x \ \ \ 0 \leq x \leq 40,000 \end{equation*}

Find the total profit function — that is, the function that descrbies the total profit the company realizes in manufacturing and selling $x$ units per month.
Find the profit when the level of production is $10,000$ units.

Solution

We obtain the total profit function by taking the difference between the total revenue realized and the total cost incurred:

\begin{equation*} \begin{split} P(x) \amp = R(x) - C(x) \\ \amp = (-0.0005x^{2} + 22x) - (-0.0001x^{2} + 12x + 10000) \\ \amp = -0.0004x^{2} + 10x - 10000 \end{split} \end{equation*}
When the production level is 10000 units per month, the company's profit is given by

\begin{equation*} P(10000) = -0.0004(10000)^{2} + 10(10000) - 10000 = 50000 \end{equation*}

or $$50,000$ per month.

Example 2.50. Revenue.

If exactly $1000$ people sign up for a charter flight, a certain travel agency charges $600$ per person. However, if more than $1000$ people sign up for the flight (assume this is the case), then each fare is reduced by $$5$ for each additional person. Letting $x$ denote the number of passengers above $1000\text{,}$ find a function giving the revenue realized by the company.

Solution

If there are $x$ passengers above $1000\text{,}$ then the number of passengers signing up for the flight in total is $1000 + x\text{.}$ The fare will be $$(600-5x)$ per passenger. Therefore, the revenue will be

\begin{equation*} \begin{split} R(x) \amp = \text{ (number of passengers) } \times \text{ (fare per passenger) } \\ \amp =(1000+x)(600-5x) \\ \amp =-5x^{2} - 4400x + 600,000 \end{split} \end{equation*}

Clearly, $x$ must be nonnegative, and $600-5x \geq 0\text{.}$ So the final revenue function is $R(x) = -5x^{2} -4400x + 600,000$ for $x$ in the interval $[0,120]\text{.}$

A company can realize a profit only if the revenue received from its customers exceeds the cost of producing and selling its products. It is therefore of interest to find out when revenue equals cost, which is referred to as the break-even point. The number of units at which revenue equals cost is the break-even quantity, the corresponding price is the break-even price, and the corresponding ordered pair gives the break-even point.

Definition 2.51. Break-Even Point.

The break-even point $(x, p)$ is defined by

\begin{equation*} (x, p)=(\text{ break-even quantity } , \text{ break-even price } ) \end{equation*}

and provides the quantity $x$ and price $p$ at which revenue equals cost.

Example 2.52. Break-Even Point.

A certain firm determines that the total cost $C(x)$ in dollars of producing and selling $x$ units is given by

\begin{equation*} C(x) = 20x + 100\text{.} \end{equation*}

Management plans to charge $$24$ per unit.

How many units must be sold for the firm to break even?
What is the profit if $100$ units of feed are sold?
How many units must be sold to produce a profit of $$900\text{?}$

Solution

The firm will break even as long as revenue just equals cost, or $R(x) = C(x)\text{.}$ From the given information,

\begin{equation*} R(x) = 24x\text{.} \end{equation*}

Substituting for $R(x)$ and $C(x)$ gives

\begin{equation*} 24x = 20x + 100\text{,} \end{equation*}

from which we get that $x = 25\text{.}$ The firm therefore breaks even by selling $25$ units (this is the break-even quantity). If the company sells more than $25$ units, it makes a profit. If it sells fewer than $25$ units, it loses money, as shown in the figure below.
First, we find the formula for the profit, $P(x)\text{.}$

\begin{equation*} \begin{split} P(x) \amp = R(x) - C(x) \\ \amp = 24x - (20x + 100) \\ \amp = 4x - 100 \end{split} \end{equation*}

Thus, $P(100) = 4(100) - 100 = 300\text{.}$ The firm will make a profit of $$300$ from the sale of $100$ units.
Let $P(x) = 900$ in the equation $P(x) = 4x-100\text{.}$

\begin{equation*} \begin{split} 900 \amp = 4x - 100 \\ 1000 \amp = 4x \\ x \amp = 250 \end{split} \end{equation*}

Sales of $250$ units will produce a profit of $$900\text{.}$

Exercises for Section 2.7.

Exercise 2.7.1.

For the following demand equations, where $x$ is the quantity demanded in units of a thousand and $p$ is the unit price in dollars, sketch the demand curve and determine the quantity demanded at the given unit price.

$p = -x^{2} + 36\text{;}$ $p = 11$
Answer
When the unit price is $$11\text{,}$ the quantity demanded is $5000\text{.}$
$p = \sqrt{9-x^{2}}$ ; $p = 2$
Answer
When the unit price is $$2\text{,}$ the quantity demanded is approx. $2236\text{.}$

Exercise 2.7.2.

For the following supply equations, where $x$ is the quantity supplied in units of a thousand and $p$ is the unit price in dollars, sketch the supply curve and determine the price at the given number of units.

$p = 2x^{2} + 18\text{;}$ $x=2000$
Answer
When the supply is $2000$ units, the price will be $26\text{.}$
$p = x^{3} + x + 10\text{;}$ $x=2000$
Answer
When the supply is $2000$ units, the price will be $20\text{.}$

Exercise 2.7.3.

For the following pairs of supply and demand equations, where $x$ represents the quantity demanded in units of a thousand and $p$ is the unit price in dollars, find the equilbrium point.

$0 = -2x^{2}-p+80$ and $0 = 15x-p+30$
Answer
$(2500,67.50)$
Solution

To find the equilbrium quantity, we can equate the supply and demand equations.

\begin{equation*} \begin{split} -2x^{2} + 80 \amp = 15x + 30 \\ -2x^{2}-15x+50 \amp = 0 \\ -(x+10)(2x-5) \amp = 0 \end{split} \end{equation*}

This gives one nonnegative solution, $x = 2.5\text{.}$ The corresponding price is

\begin{equation*} p = 15(2.5)+30 = 67.5\text{.} \end{equation*}

This tells us that the equilibrium quantity is $2500\text{,}$ and the equilibrium price is $$67.50\text{.}$
$11p + 3x = 66$ and $2p^{2} + p - x = 10$
Answer
$(11000,3)$

Exercise 2.7.4.

A production company has a monthly fixed cost of $$100,000$ and a variable cost of $$14$ for each unit produced. The commodity sells for $$20$ per unit.

What is the cost function?
Answer
$C(x) = 14x + 100,000$
What is the revenue function?
Answer
$R(x) = 12x$
What is the profit function?
Answer
$P(x) = 6x - 100,000$
Compute the profit (or loss) corresponding to manufacturing levels of $12,000$ and $20,000$ units.
Answer
Loss of $$28,000\text{.}$ Profit of $$20,000$

Exercise 2.7.5.

The price of ivory is determined by a variety of legal market sources and illegal ivory trades. The World Wildlife Fund, an international non-governmental organization trying to preserve wilderness, determined that the ivory price can be approximated by the piecewise-defined function

\begin{equation*} p(t) = \begin{cases} 8.37t + 7.44 \amp \text{ if } 0 \leq t \leq 8 \\ 2.84t + 51.68 \amp \text{ if } 8 \lt t \leq 30 \end{cases} \end{equation*}

where $t$ is measured in years, with $t = 0$ corresponding to the beginning of the year $1970\text{,}$ and $p(t)$ is measured in dollars per kilogram.

Sketch the graph of $p(t)\text{.}$
Answer
What was the price of ivory at the beginning of $1970\text{?}$
Answer
$$7.44$ per kg
Solution
The beginning of $1970$ corresponds to $t = 0\text{.}$ Therefore, the price of ivory was $p(0) =$ $ 7.44 per kg.
What was the price of ivory at the beginning of $1990\text{?}$
Answer
$$108.48$ per kg
Solution
The beginning of $1990$ corresponds to $t = 20\text{.}$ Looking at the definition of $p\text{,}$ we see that $p(20) = 2.84(20)+51.68 =$ $108.48 per kg.

Exercise 2.7.6.

In the accompanying figure, $L_{1}$ is the demand curve for the version $X$ of cell phones, and $L_{2}$ is the demand curve for the version $Y$ of cell phones. Which line has greater slope? Interpret your results.

Solution

The slope of $L_{2}$ is greater than the slope of $L_{1}\text{.}$ For each drop of a dollar in price, the quantity demanded for version $X$ is greater than for version $Y\text{.}$

Exercise 2.7.7.

The demand function for a certain product is given by

\begin{equation*} p = \dfrac{3000}{2x^{2} + 100} \ \ \ (0 \leq x \leq 10) \end{equation*}

where $x$ (measured in units of a thousand) is the quantity demanded per week and $p$ is the unit price in dollars. Sketch the graph of the demand function. What is the price that corresponds to a quantity demanded of $10,000$ units?

Answer

$p = 10$

Exercise 2.7.8.

A certain manufacturer has determined that the weekly demand and supply functions for their product is given by

\begin{equation*} p(q) = 144-q^{2} \end{equation*}

\begin{equation*} p(q) = 48 + \frac{1}{2}q^{2} \end{equation*}

respectively, where $p$ is measured in dollars and $q$ is measured in units of a thousand. Find the equilibrium point.

Answer

$(8000,80)$

Exercise 2.7.9.

Suppose that the demand and price for a certain model of wristwatch are related by

\begin{equation*} p = D(q) = 16-1.25q \end{equation*}

where $p$ is the price in dollars and $q$ is the demand in units of hundreds. The price and supply of the watch are related by

\begin{equation*} p = S(q) = 0.75q \end{equation*}

where $p$ is the price in dollars and $q$ is the supply of watches in units of hundreds.

Find the price when the demand is $0$ watches
Answer
$$16$
$400$ watches
Answer
$$11$
$800$ watches
Answer
$$6$
Find the demand when the price is $$8$
Answer
$640$ watches
$$10$
Answer
$480$ watches
$$12$
Answer
$320$ watches
Sketch $D(q)$
Answer
Find the supply when the price is $$0$
Answer
$0$ watches
$$10$
Answer
Approximately $1333$ watches
$$20$
Answer
Approximately $2667$ watches
Sketch $S(q)$ on the same plot as for part (g)
Answer
Find the equilibrium quantity and the equilibrium price
Answer
$(8,6)$

Exercise 2.7.10.

Joanne sells T-shirts at community festivals and creaft fairs. Her marginal cost to produce one T-shirt is $$3.50\text{.}$ Her total cost to produce $60$ T-shirts is $$300\text{,}$ and she sells them for $$9$ each.

Find the linear cost function for Joanne's T-shirt production.
Answer
$C(x) = 3.50x + 90$
How many T-shirts must she produce and sell in order to break even?
Answer
$17$ T-shirts
How many T-shirts must she produce and sell to make a profit of $$500\text{?}$
Answer
$108$ T-shirts

Exercise 2.7.11.

You are the manager of a firm. You are considering the manufacture of a new product, so you ask the accounting department for cost estimates and the sales deparment for sales estimates. After you recieve the data, you must decide whether to go ahead with production of the new product. Analyze the given data (find a break-even quantity) and then decide what you would do in each case. Include the profit function.

$C(x) = 105x + 6000\text{;}$ $R(x) = 250x\text{;}$ no more than $400$ units can be sold.
Answer
Break-even quantity is about $41$ units; produce. $P(x) = 145x - 6000\text{.}$
Solution

We find the break-even point by setting $C(x) = R(x)\text{:}$

\begin{equation*} \begin{split} 105x+6000 \amp = 250x \\ 145x \amp = 6000\\ x \amp \approx 41 \end{split} \end{equation*}

The corresponding profit function is

\begin{equation*} P(x) = R(x) - C(x) = 145x - 6000 \ \ \ 0 \leq x \leq 400\text{.} \end{equation*}

We would therefore decide to produce this product, since there is opportunity to profit.
$C(x) = 1000x + 5000\text{;}$ $R(x) = 900x\text{.}$
Answer
Break-even quantity is $-50$ units; don't produce. Impossible to make a profit when $C(x) > R(x)$ for all $x > 0\text{.}$ $P(x) = -100x - 5000$ (always a loss).
Solution

Let's first find the profit function.

\begin{equation*} \begin{split} P(x) \amp = 900x - (1000x + 5000) \\ \amp = -100x - 5000 \\ \amp = -(100x+5000) \end{split} \end{equation*}

The break even point corresponds to $P(x) = 0 \implies x = -50\text{.}$ We can see that, for all $x \geq 0\text{,}$ $P(x) \leq 0\text{.}$ Therefore, we would decide to not produce.