Solutions to Quadratic Equations Word Problems
Area Problems:
9. A local building code requires that all factories must be surrounded by a lawn. The width of
the lawn must be uniform and the area must equal the area of the factory. What must be the
width of the lawn surrounding a rectangular factory that measures 120 m by 80 m?
Information given:
Area factory = 120m x 80m
Area lawn = Area factory
2
Therefore, Area lawn = Area factory = 120m x 80 m = 9600 m
Unknown information: width of the lawn
120
x
80
Let the width of the lawn be x
Area lawn = Area total - Area lawn
9600 = (120 + 2x)( 80 + 2x) – 9600
2
9600 = (9600 + 240x +160x + 4x ) -9600
2
9600 = 400x + 4x
2
4x + 400x -9600 = 0
2
4(x + 100x -2400) = 0
4(x +120)(x -20) = 0
Therefore (x + 120) = 0 or (x – 20) = 0
Therefore x = -120 or x = 20
Since x is the width, -120 is inadmissible, therefore x = 20
Therefore, the width of the lawn is 20 m
10. A factory is to be built on a lot that measures 80m x 60m. A lawn of uniform width and equal
in area to the factory must surround the factory. What dimensions must the factory have?
Information given:
2
Area lot = Area lawn + Area factory = 80m x 60m = 4800 m
Area lawn = Area factory
Unknown information: dimensions of the factory
x
Let the width of the lawn be x
Area factory = Area lot - Area lawn
(80 – 2x)(60 - 2x) = 4800 - (80 – 2x)(60 - 2x)
2(80 – 2x)(60 - 2x) = 4800
2
2(4800 -160x -120x + 4x ) = 4800
2
9600 -560x +8x = 4800
2
8x -560x + 9600 – 4800 = 0
2
8x -560x + 4800 = 0
2
8(x - 70x +600) = 0
8(x - 60) (x – 10) = 0
Therefore (x - 60) = 0 or (x – 10) = 0
Therefore x = 60 or x = 10
Since x represents the width, x= 60 is inadmissible, therefore x = 10
The dimensions of the factory are (80 - 2(10))x (60 - 2(10)) = 60 x 40
Therefore, the dimensions of the factory are 60m x 40m
60
80
12. A matte of uniform width is to be placed around a painting so that the area of the matted
surface is twice the area of the picture. If the outside dimensions of the matte are 40 cm and 60
cm, find the width of the matte.
Information given:
Area matte = 2 Area picture
2
Area total = 40 cm x 60 cm = 2400 cm
Unknown information: width of the matte
x
40
Let the width of the matte be x
Area matte = Area total - Area picture
2(Area picture) = Area total - Area picture
3(Area picture) = Area total
3(60 – 2x) (40 – 2x) = 60 x 40
2
3(2400 – 200x + 4x ) = 2400
2
7200 – 600x + 12x = 2400
2
12x -600x +4800 = 0
2
12(x -50x +400) = 0
24(x – 40)( x – 10) = 0
60
Therefore (x - 40) = 0 or (x – 10) = 0
Therefore x = 40 or x = 10
Since x is the width, x = 40 is inadmissible, therefore x = 10
Therefore, the width of the matte is 10 cm
14. A picture 20cm wide by 10cm high is to be centrally mounted on a rectangular frame with
total area three times the area of the picture. Assuming equal margins for all four sides, find the
width of the margin.
Information given:
Area total = 3 Area picture
2
Area picture = 20 cm x 10 cm = 200 cm
Unknown information: width of the margin
x
Let the width of the margin be x
3(Area picture) = Area total
3(20)(10) = (20 + 2x)(10 + 2x)
2
600 = (200 + 40x + 20x + 4x )
2
600 = 200 +60x + 4x
2
4x + 60x – 400 = 0
2
4(x +15x -100) = 0
4(x + 20)(x – 5) = 0
Therefore (x + 20) = 0 or (x – 5) = 0
Therefore x = -20 or x = 5
Since x is the width, x = -20 is inadmissible, therefore x = 5
Therefore, the width of the margin is 5 cm
10
20
Merchandising Problems:
21. The Puck Heads hockey team averages 15000 people at each hockey game with the average
price of a ticket being $60. In the championship game, admission prices will increase. For
every $5 increase in ticket price, 1000 fewer people will attend. What price will maximize the
Puck Heads’ revenue?
Let x be the number of times the price is increased by $5.
Price per ticket = (60 + 5x)
Tickets Sold = (15000 -1000x)
Revenue = Price x Tickets Sold
Revenue = (60 + 5x) (15000 - 1000x)
To find the maximum revenue, you must find the vertex. To do this you must first find the zeros.
To find the zeros, set the revenue to 0
0 = (60 + 5x) (15000 - 1000x)
Therefore (60 + 5x) = 0 and (15000 - 1000x) = 0
Therefore x = -12 and x = 15
The x co-ordinate for maximum revenue is x = (-12 + 15)/2 = 3/2
The new price is (60 + 5 (3/2)) = 67.50
Therefore, the price to maximize the revenue is $67.50
22. The local Transit commission’s single fare price is $1.20 cash. On a typical day, 240,000
people take the transit and pay the single-fare price. To reflect higher costs, single fare prices
will be increased but surveys show that every $0.05 increase in fare will reduce ridership by
8000 riders daily. What single-fare price will maximize revenue for the commission based on
single fares?
Let x be the number of times the price is increased by $0.05.
Price per ticket = (1.20 + 0.05x)
Tickets Sold = (240000 -8000x)
Revenue = Price x Tickets Sold
Revenue = (1.20 + 0.05x) (240000 - 8000x)
To find the maximum revenue, you must find the vertex. To do this you must first find the zeros.
To find the zeros, set the revenue to 0
0 = (1.20 + 0.05x) (240000 - 8000x)
Therefore (1.20 + 0.05x) = 0 and (240000 - 8000x) = 0
Therefore x = -24 and x = 30
The x co-ordinate for maximum revenue is x = (-24 + 30)/2 = 3
The new price is (1.20 + 0.05(3)) = 1.35
Therefore, the price to maximize the revenue is $1.35
23. A company sells 9000 pairs of slacks a month at an average of $70 each. The store is going
to increase prices in order to increase profits. Sales forecasts indicated that sales will drop by
200 for every dollar increase in price. On average, the company pays $30 for each pair of slacks
that it sells. What price will maximize the profits?
Let x be the number of times the price is increased by $1.
Price per pair of slacks = (70 + x)
Pants Sold = (9000 - 200x)
Revenue = Price x Slacks Sold
Revenue = (70 + x) (9000 - 200x)
Cost
= 30(9000 – 200x)
To find the maximum profit: Profit = Revenue – Cost
P = (70 + x)(9000 – 200x) – 30(9000 – 200x)
= (9000 – 200x)(70 + x – 30)
= (9000 – 200x)(40 + x)
Set P = 0
Therefore
0 = (9000 – 200x)(40 + x)
9000 – 200x = 0
X = 45
or 40 + x = 0
x = -40
The x – co-ordinate for maximum profit is:
x
45  40 5

2
2
New Price = 70 + 2.5 = $72.5
Therefore, the price to maximize the profits is $72.50
24. A hardware store currently sells light bulbs each week at a price of $7.00 per package. To
increase sales and reach more customers, the store decides to reduce the price of the package
knowing that every $0.10 decrease in price will result in 5 more sales. What price will maximize
total revenue?
Let x be the number of times the price is decreased by $0.10.
Price per light bulb package = (7 – 0.1 x)
Light Bulbs Sold = (300 + 5x)
Revenue = Price x Light Bulbs Sold
Revenue = (7 – 0.1x) (300 + 5x)
To find the maximum revenue, you must find the vertex. To do this you must first find the zeros.
To find the zeros, set the revenue to 0
0 = (7 – 0.1x) (300 + 5x)
Therefore (7 - 0.1x) = 0 and (300 + 5x) = 0
Therefore x = 70 and x = - 60
The x co-ordinate maximum revenue is x = (70 - 60)/2 = 5
The new price is (7 – 0.1 (5)) = 6.50
Therefore, the price to maximize the revenue is $6.50