What are Linear Cost Functions?
By Robert O
Explain linear cost functions with examples:
There is nothing unique about linear cost functions if you are already familiar with general linear functions, which are of the form y = m x+c The equation is not new to you, right?
When it comes to linear cost functions, we will be dealing with the cost of a commodity or service that has two parts. To get what it will cost to purchase the product or receive the service, you need to consider both the variable cost and a fixed cost. The equation is of the form:
![\[Y=A x+B\]](data:image/svg+xml;nitro-empty-id=MzAwOjEzNDg=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTcgMTUiIHdpZHRoPSI5NyIgaGVpZ2h0PSIxNSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Where:
- Y is the total cost
- A is the cost per unit, which depends on the number of units x.
- B is the fixed cost or fixed charge.
Think of your utility bills as an example. There is usually a fixed charge that does not depend on the units that you use. So, whether you used energy in a month or not, you have to pay the fixed charges.
How do we solve linear cost problems?
The key to solving problems involving linear cost is to understand the question statement. You calculate the value of B and A using the information from it. With these constants known, you can calculate the total cost y for each input variable x. Let’s understand it by using examples.
Example 1
The total cost of producing two dresses is 130 dollars, and the production cost of 5 similar dresses is 190 dollars. By assuming a linear cost function, what is the cost of producing 8 such dresses?
Solution
We use the linear cost function to form two linear equations and then solve for the unknown.
From the general equation:
![\[Y=A x+B\]](data:image/svg+xml;nitro-empty-id=MzAwOjMwMjM=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTcgMTUiIHdpZHRoPSI5NyIgaGVpZ2h0PSIxNSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[130=2 A+B, \text { total cost of producing } 2 \text { dresses is } 130 .\]](data:image/svg+xml;nitro-empty-id=MzAwOjM1MTc=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDI2IDE3IiB3aWR0aD0iNDI2IiBoZWlnaHQ9IjE3IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[190=5 A+B, \text { total cost of producing } 5 \text { dresses is } 190 .\]](data:image/svg+xml;nitro-empty-id=MzAwOjQwMTE=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDI2IDE3IiB3aWR0aD0iNDI2IiBoZWlnaHQ9IjE3IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
It is now clear that we have a system of two linear equations. We can solve it by Gaussian elimination, algebraic substitution, or graphical method. In this case, we will use algebraic substitution.
Using the first equation:
![\[130=2 A+B\]](data:image/svg+xml;nitro-empty-id=MzAwOjQ2Nzg=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTA3IDE1IiB3aWR0aD0iMTA3IiBoZWlnaHQ9IjE1IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[B=130-2 A, \text { substitute this in the second equation. }\]](data:image/svg+xml;nitro-empty-id=MzAwOjUxNTk=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDE5IDE3IiB3aWR0aD0iNDE5IiBoZWlnaHQ9IjE3IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[190=5 A+130-2 A\]](data:image/svg+xml;nitro-empty-id=MzAwOjU1OTU=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTYzIDE1IiB3aWR0aD0iMTYzIiBoZWlnaHQ9IjE1IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[A=\frac{60}{3}=20\]](data:image/svg+xml;nitro-empty-id=MzAwOjYwMzM=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTAwIDM2IiB3aWR0aD0iMTAwIiBoZWlnaHQ9IjM2IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
Now, substitute the value of A in any equation to find B.
![\[B=130-2 A\]](data:image/svg+xml;nitro-empty-id=MzAwOjY1Mjc=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTA4IDEzIiB3aWR0aD0iMTA4IiBoZWlnaHQ9IjEzIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[B=130-2(20)\]](data:image/svg+xml;nitro-empty-id=MzAwOjY5NTk=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTI2IDE5IiB3aWR0aD0iMTI2IiBoZWlnaHQ9IjE5IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[B=90\]](data:image/svg+xml;nitro-empty-id=MzAwOjczODM=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNTYgMTIiIHdpZHRoPSI1NiIgaGVpZ2h0PSIxMiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Form a linear cost function:
![\[Y=20 x+90\]](data:image/svg+xml;nitro-empty-id=MzAwOjc4NDg=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTA2IDE0IiB3aWR0aD0iMTA2IiBoZWlnaHQ9IjE0IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
The total cost of producing 8 dresses is:
![\[Y=20(8)+90=250\]](data:image/svg+xml;nitro-empty-id=MzAwOjgzMzE=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTY4IDE5IiB3aWR0aD0iMTY4IiBoZWlnaHQ9IjE5IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
The total cost of producing 8 dresses is 250 dollars.
What if we graph our linear cost function
![\[Y=20 x+90 ?\]](data:image/svg+xml;nitro-empty-id=MzAwOjg4NjU=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTEzIDE0IiB3aWR0aD0iMTEzIiBoZWlnaHQ9IjE0IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
Graph generated from https://www.desmos.com/
The line crosses the y-axis (y-intercept) at the fixed cost point. The value is 90. The x-axis represents the units produced, and the y-axis represents the total cost.
Example 2
A fixed cost of running machine A is 75 dollars and a variable cost of 3 dollars for every unit of product from this machine. Another machine, B, has a fixed cost of 60 dollars and a variable cost for producing an item of 4.5 dollars. What number of items will make the running cost of both machines the same?
Solution
Let us use the information from the problem statement to create two linear equations.
![\[Y=A x+B\]](data:image/svg+xml;nitro-empty-id=MzA0OjEyMDQ=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTcgMTUiIHdpZHRoPSI5NyIgaGVpZ2h0PSIxNSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[Y=3 x+75\]](data:image/svg+xml;nitro-empty-id=MzA0OjE2MzI=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTUgMTUiIHdpZHRoPSI5NSIgaGVpZ2h0PSIxNSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[Y=4.5 x+60\]](data:image/svg+xml;nitro-empty-id=MzA0OjIwNjM=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTEwIDE1IiB3aWR0aD0iMTEwIiBoZWlnaHQ9IjE1IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
We solve for x, which is the value we need, by equating the right side of the equations.
![\[3 x+75=4.5 x+60\]](data:image/svg+xml;nitro-empty-id=MzA0OjI1OTQ=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTU0IDE1IiB3aWR0aD0iMTU0IiBoZWlnaHQ9IjE1IiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[-1.5 x=-15\]](data:image/svg+xml;nitro-empty-id=MzA0OjMwMjU=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTAwIDEzIiB3aWR0aD0iMTAwIiBoZWlnaHQ9IjEzIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[x=10\]](data:image/svg+xml;nitro-empty-id=MzA0OjM0NDk=-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNTIgMTIiIHdpZHRoPSI1MiIgaGVpZ2h0PSIxMiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
The cost of running the two machines is the same when we produce the 10th item.
Remarks
The step to solving linear cost functions is to understand the statement to form a system of two linear equations. We then solve to find the unknown values, after which we can compute the cost of producing any unit. You only need to be keen on interpreting the problem statement.
About the Author
This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.