Today's soundtrack is Christmas With the Mills Family, a Spotify playlist that includes my favourite Christmas songs.
This evening, I'm working on the final chapter of this unit, "Modelling and Solving Problems With Quadratic Functions." In this chapter, I'll be going over different ways that I can write out a quadratic function to model and solve a problem.
In general, when faced with a word problem involving quadratic functions, we should employ the following method:
Sketch a diagram, if applicable
Define variables for information that we don't yet know
Use the given information to write an equation with a single variable
Solve the equation for the appropriate variable, converting to vertex form if needed
In questions that ask us to find the minimum or maximum product of two numbers based only on their difference, we need to turn it into an equation with only one variable and solve through substitution. For example, if we know that the sum of the length and width of fencing materials is 20km, and we are asked to determine the dimensions of a plot of land that would that produce the maximum area, we would do the following:
Set up variables
let L = length; let w = width; let a = area
Create equations
We know that the sum of the two sides is twenty, giving us the equation L+w=20
The product of the dimensions will give us our maximum area, so L ⋅ w = a.
Convert equations to make them useable for substitution with a single variable
Rather than saying L + w = 20, we can say L = 20 - w
Using substitution, insert the known values into our equation
Since we know that L = 20 - w, we can substitute that value into our equation, giving us a = (20-w)w
Expand the equation
Apply distributive properties to a = (20-w)w to get -w² + 20w
Complete the square
Half of 20 is 10; 10² is 100, so we now have -w² + 20w + 100 - 100.
-w² + 20w + 100 is a perfect square; our equation is now a = -(w+10). We still need to add that "-100" back on, but since we popped the negative sign outside of the brackets, to retain its value, we need to invert its symbol, making our updated equation a = -(w+10) + 100.
Find the answer
Our two variables that would give us the dimensions were L and w. When a quadratic equation is written out in the vertex form as above, the x-coordinate of the vertex is the second term in the brackets, which is w = 10. We determined in step 3 that L = 20 - w. Using substitution, we calculate that as L = 20 - 10; therefore, L = 10. So both L and w = 10.
The dimensions that would give us the maximum area is 10km length and 10km width.
I found this YouTube video very helpful while working on this lesson.
The example given above was adapted from the material on page 323 in Pearson's Pre-Calculus 11 Worktext.