2.4. Applications
Applications
Now that we have learned to determine equations of lines, we can apply these ideas to real-life equations.
Example 2.4.1
In the above problem, $0.50 per mile is referred to as the variable cost, and the flat charge $5 as the fixed cost. Now if we look at our cost equation y = .50x + 5, we can see that the variable cost corresponds to the slope and the fixed cost to the y-intercept.
Example 2.4.2
The variable cost to manufacture a product is $10 and the fixed cost $2500. If x represents the number of items manufactured and y the total cost, write the cost function.
Solution
The variable cost represents the slope and the fixed cost represents the y-intercept. Therefore, m = 10 and y = 2500.
The cost equation is y = 10x + 2500.
Example 2.4.3
It costs $750 to manufacture 25 items, and $1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.
Solution
We let x = the number of items manufactured, and let y = the cost.
Solving this problem is equivalent to finding an equation of a line that passes through the points (25, 750) and (50, 1000).
Therefore, the partial equation is y = 10x + b.
By substituting one of the points in the equation, we get b = 500.
Therefore, the cost equation is y = 10x + 500.
Now to find the cost of 100 items, we substitute x = 100 in the equation y = 10x + 500.
So the cost = y = 10(100) + 500 = 1500.
Example 2.4.4
Let us look at what is given.
| Centigrade | Fahrenheit |
| 0 | 32 |
| 100 | 212 |
Again, solving this problem is equivalent to finding an equation of a line that passes through the points (0, 32) and (100, 212).
Since we are finding a linear relationship, we are looking for an equation y = mx + b, or in this case F = mC+ b, where x or C represent the temperature in Celsius, and y or F the temperature in Fahrenheit.
slope
The equation is
Substituting the point (0, 32), we get
Now to convert 30 degrees Celsius into Fahrenheit, we substitute C = 30 in the equation
Example 2.4.5
The population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.
Solution
The problem can be made easier by using 1970 as the base year, that is, we choose the year 1970 as the year zero. This will mean that the year 1986 will correspond to year 16, and the year 2010 as the year 40.
Now we look at the information we have.
Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 18) and (16, 26).
The equation is .
Substituting the point (0, 18), we get:
Now to find the population in the year 2010, we let x = 40 in the equation:
So the population of Canada in the year 2010 is estimated as 38 million.
| Year | Population |
| 0 (1970) | 18 million |
| 16 (1986) | 26 million |
More Applications
In this section, you will learn to:
- Solve a linear system in two variables.
- Find the equilibrium point when a demand and a supply equation are given.
- Find the break-even point when the revenue and the cost functions are given.
In this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.
Example 2.4.6
Find the intersection of the line y = 3x − 1 and the line y = −x + 7.
Solution
We graph both lines on the same axes, as shown below, and read the solution (2, 5).
Finding the intersection of two lines graphically is not always easy or practical; therefore, we will now learn to solve these problems algebraically.
At the point where two lines intersect, the x and y values for both lines are the same. So in order to find the intersection, we either let the x-values or the y-values equal.
If we were to solve the above example algebraically, it will be easier to let the y-values equal. Since y = 3x − 1 for the first line, and y = −x + 7 for the second line, by letting the y-values equal, we get:
By substituting x = 2 in any of the two equations, we obtain y = 5. Hence, the solution (2, 5).
One common algebraic method used in solving systems of equations is called the elimination method. The object of this method is to eliminate one of the two variables by adding the left and right sides of the equations together. Once one variable is eliminated, we get an equation that has only one variable for which it can be solved. Finally, by substituting the value of the variable that has been found in one of the original equations, we get the value of the other variable. The method is demonstrated in the example below.
Example 2.4.7
Example 2.4.8
Solve the system of equations x + 2y = 3 and 2x + 3y = 4 by the elimination method.
Solution
If we add the two equations, none of the variables are eliminated. But the variable x can be eliminated by multiplying the first equation by -2, and leaving the second equation unchanged.
Substituting y = 2 in x + 2y = 3, we get
Therefore, the solution is (-1, 2).
Example 2.4.9
Solve the system of equations 3x − 4y = 5 and 4x − 5y = 6.
Solution
Supply, Demand and the Equilibrium Market Price
In a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.
Example 2.4.10
Break-Even Point
Example 2.4.11
Practice questions
1. The variable cost to manufacture an item is $20, and it costs a total of $750 to produce 20 items. If x represents the number of items manufactured and y the cost, write the cost function.
2. A person who weighs 150 pounds has 60 pounds of muscles, and a person that weighs 180 pounds has 72 pounds of muscles. If x represents the body weight and y the muscle weight, write an equation describing their relationship. Use this relationship to determine the muscle weight of a person that weighs 170 pounds.
3. In 2005, an average house in Greater Toronto Area cost $335,907 and the average house in 2018 cost $787,300. Assuming a linear relationship, predict the price of a similar house in the year 2025.
4. In 2010 there were 11,386 laboratory-confirmed cases of gonorrhea reported in Canada. In 2015, the number of cases increased to 19,845. Assuming a linear relationship, how many cases of gonorrhea might we expect in 2030?
5. The supply curve for a product is y = 2000x + 13000, and the demand curve is y = -1000x + 28000, where x represents the price and y the number of items. At what price will the supply equal demand, and how many items will be produced at that price?
6. A company that produces toys has a fixed cost of $10,725, and variable cost of 20 cents a toy. Find the break-even point if the toys sell for $1.50 each.