Today in class we focused on Quadratic Word Problems. There are three types of the word problems.

We did the first kind of word problem:

1.) One thing will increase invalue while another thing will decrease.

Ex. Computer software sells for $20.00 each, 300 people will buy. For every $5 increase in price 30 less people will buy. What is maximum revenue?


Revenue = Price x number sold

Price -> 20 + 5x
number sold -> 300 - 30x

Revenue = Price x number sold
Revenue = (20 + 5x)(300 - 30x) <---fill in the variables
Revenue = 6000 - 600x + 1500x - 150x² <--- factor it out
Revenue = 6000 + 900x - 150x²
Revenue = -150x² + 900x + 6000 <---put in correct order
Revenue = -150(x² - 6x + ___) + 6000 <--- factor out -150
^to find the value of the 3rd term you would use the complete square method
Revenue = -150(x² - 6x + 9 ) + 6000 + 1350 <--- again use complete square method to balance out the equation
Revenue = -150(x - 3)² + 7350 <--- the result in completing the square

Vertex = ( 3, 7350)

Price = 20 + 5x
Price = 20 + 5(3)
Price = 20 + 15
Price = 35

Number sold = 300 - 30x
Number sold = 300 - 30(3)
Number sold = 300 - 90
Number sold = 210

Revenue = Price x Number sold
Revenue = 35 x 210
Revenue = 7350

The second type of word problem:

2.) Finding an enclosed area.

ex. What is the maximum rectangular area that can be enclosed by 120m of fencing if one side of the rectangle is an existing wall?

Area = length x width
Area = (120 - 2w)w
Area = 120w - 2w²
Area = -2w² + 120 <--rearrange in correct order
Area = -2(w² - 60w + __ ) <--- complete square
Area = -2(w² -60w + 900) + 1800
Area = -2(w - 30)² + 1800

Vertex = (30, 1800)

Width = 30
Length = 120 - 2x
Length = 120 -2(30)
Length = 120 - 60
Length = 60

Maximum Area = 1800 <-- max area is always the y- value
check by using the formula A=LxW

A=LxW
A= 60 x 30
A= 1800


The third type of word equation :

3.) Given an equation.

ex. A projectile is shot straight up from a height of 6m with initial velocity of 80 m/s. It's height is given by the equation :

h = 6 + 80t - 5t²

After how many seconds will it reach it's maximum height? What is that height?

Maximum height = y value of the vertex.

h = 6 + 80t - 5t²

h = -5t² + 80t + 6 <-- rearrange in correct order

h = -5 (t² - 16t + __) + 6 <--- complete square

h = -5(t² - 16t + 64) + 6 +320

h = -5 (t - 8)² + 326

Vertex = (8,326)

Maximum height = 326

Seconds = 8

Well that's what we focused on for today. The next scribe is Kirstie.