Quadratic Word Problems

Quadratic Word ProblemsMax/min

Finding Roots

ProbeBelow are 6 different representations of 3 different quadratic relationships. Pair them up, and name the 6 forms.

)13,5(),( yxyx 2)2()9( xy 1032 xxy

x y-8 -26-7 -11-6 -2-5 1-4 -2 -3 -11-2 -26)5)(2( xxy

ProbeBelow are 6 different representations of 3 different quadratic relationships. Pair them up, and name the 6 forms.

)13,5(),( yxyx 2)2()9( xy 1032 xxy

x y-8 -26-7 -11-6 -2-5 1-4 -2 -3 -11-2 -26)5)(2( xxy

mapping rule equation in general formequation in transformational form

equation in factored formtable of values

graphical (parabola)

You should know the following about quadratic functions:

How to graph themHow to find the vertexHow to find the x- and y-interceptsHow to find the equation from the patternHow to find the equation from the graphHow to change from one form to another

Recap

Quadratic Word ProblemsThere are basically two types of quadratic word problems:

Those that ask you to find the vertex

Those that ask you to find the roots

By giving you the equation

By giving you the information to find the equation

Those that ask you to find the vertex

Those that ask you to find the roots

(…these are the harder

ones)

Max/Min Values

abf

ab

2,

2

y-value has a min when a > 0

y-value has a max when a < 0

We have seen that the vertex of a quadratic in general form (y = ax2 + bx + c) is given by:

The y-value of the vertex is either the maximum value the function can have or it’s the minimum value the function can have.

808.0

64)4.0(2

642

p

p

p

abp

Example 1 (equation given)

A small business’ profits over the last year have been related to the price of the its only product. The relationship isR(p) = −0.4p2 + 64p − 2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars.

a. What price would maximize the revenue?

Thus, a price of $80 would maximize the revenue.

The word maximize screams:“FIND THE VERTEX!!”



b. What is the maximum revenue possible?

The maximum revenue is $160 000.

The answer to this question is the y-value (R-value) of the vertex. 160)80(

240051202560)80(24005120)6400(4.0)80(2400)80(64)80(4.0)80(

802

RRRRp



c. How much money would they lose if they gave the product away?

The business would lose $2 400 000.

This question is talking about a price of 0 or p = 0

2400)0(2400)0(64)0(4.0)0(

02

RRp

Max/Min Word ProblemsExample 2: (no equation given)

A farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the total enclosed area. According to regulations the pens need to find these relative dimensions. What is the maximum area he can enclose, and what are its dimensions?

sheep pen

pig pen

kids pen

Again the word MAXIMUM means VERTEX…but…

we don’t have an equation!

Hint: Start with the value you are trying to maximize/minimize



sheep pen

pig pen

kids pen

Using the hint we get:Area of a rectangle:A = (3w)(8L)A = 24wL

But there are too many variables!(A, w and L).

Hint: Find an equation using both L and w, and use substitution to eliminate one of them

The total fencing is 800m. This means 800 = 3w+3w+3w+8L+8L+w+2L800 = 10w + 18Lorw = 80 – 1.8L Sub it in!



sheep pen

pig pen

kids pen

This looks very promising because there are only two variables (A and L) and it is a quadratic (there is an L2)

Now find the vertex! (Because it is asking for MAXimum area)

Area of a rectangle:A = 24wLA = 24(80 – 1.8L)LA = 1920L – 43.2L2

22.22)2.43(2)1920(

2

L

L

abL Now we know

that the maximum area occurs whenL = 22.22m.



sheep pen

pig pen

kids pen

A = 1920L – 43.2L2

Now we can find this maximum area:

Maximum area:A = 1920(22.22) - 43.2(22.22)2

A = 21 333m2

Now we can also find the dimensions:

L = 22.22mw = 80 – 1.8L = 80 – 1.8(22.22) = 40m

w = 80 – 1.8L

Your Turn

2w + L = 75A = wLL = 75 – 2w

75.18)2(2

752

w

w

abw

A lifeguard has 75m of rope to section off the supervised area of the beach. What is the largest rectangular swimming area possible?

wL

wA = w(75 – 2w)A = 75w – 2w2

Maximum area:A = 75(18.75) – 2(18.75)2 A =1406.25 – 703.125A = 703.125m2

Roots of a QuadraticWe have 3 ways to find the roots of (solve) a quadratic equation:

1. Factoring(sometimes fast, sometimes not possible)

2. Completing the square(always possible, takes too long)

3. Using the quadratic root formula(always possible, pretty quick) a

acbbx2

42

Roots Word ProblemsExample 3 (equation given)

A duck dives under water and its path is described by the quadratic function y = 2x2 – 4x, where y represents the position of the duck in metres and x represents the time in seconds.

The duck is no longer underwater when the depth is 0. We can plug in y = 0 and solve for x.

)2(20420 2

xx

xx

020

xx

220

xx

So x = 0 or 2

a. How long was the duck underwater?

The duck was underwater for 2 – 0, or 2 seconds

We can plug in y = -5 and solve for x.

b. When was the duck at a depth of 5m?

5420

4252

2

xxxx

4244

440164

)2(2)5)(2(4)4()4(

24

2

2

x

x

x

aacbbx

We cannot solve this because there’s a negative number under the square root.We conclude that the duck is never 5m below the water.Let’s check by finding the minimum value of y.

1)2(2)4(

2

abx

2)1(4)1(2 2

yy



Roots Word Problems

We can plug iny = -0.5 and solve for x. This will give us the times when the duck is at exactly 0.5m below.

c. How long was the duck at least 0.5m below the water’s surface?

5.0420

425.02

2

xxxx

sorx

x

x

x

x

aacbbx

87.114.04

46.344

1244

4164

)2(2)5.0)(2(4)4()4(

24

2

2

The duck was exactly 0.5m below at x = 0.14s and at x = 1.87s. Therefore it was below 0.5m for 1.73s



Roots Word Problems

Roots Word ProblemsExample 4: (no equation given)

A rectangular lawn measures 8m by 6m. The homeowner mows a strip of uniform width around the lawn, as shown. If 40% of the lawn remains un-mowed, what is the width of the strip?

-X-

-X-

40%

The dimensions of this un-mowed rectangle are: 8 – 2x and 6 – 2x.

-----8 – 2x----- -6 –

2x-

40% of this is unmowed: 48 × 0.40 = 19.2

The total area of the lawn is 8 × 6 = 48

So 19.2 = (8 – 2x)(6 – 2x)

8.282840

2.19482840

482842.19

41216482.19

2

2

2

2

xxxxxx

xxx

This is a quadratic equation! What a perfect time to use the Quadratic Root Formula to solve for x.

mormx

x

x

aacbbx

xx

75.525.1

898.1728

)4(2)8.28)(4(4)28()28(

24

8.282840

2

2

2

The mowed strip has a width of 1.25m



40%

-----8 – 2x-----

-6 –

2x-

Let’s check:

If x = 1.25m then the

length is 8 – 2x= 8 – 2(1.25)= 5.5m

width is 6 – 2x= 6 – 2(1.25)= 3.5m

A = (5.5)(3.5)A =19.2

Which was 40% of the total area!

40%

-----8 – 2x-----

-6 –

2x-




Since there is no diagram, we have to define our variablesThe question indicates two equations relating these two variables.Again, we have a substitution situation (like example 2). Solve the simpler equation for a variable and plug it in to the other equation.We can turn this into a quadratic equation in general form. Let’s do it because then we can use the Quadratic Root Formula to solve for g.Let’s go!

Two numbers have a difference of 18. The sum of their squares is 194.What are the numbers?

Let s represent one of the numbers Let g represent the other

s – g = 18s2 + g2 = 194

s = 18 +g

(18 + g)2 + g2 = 194324 + 36g + g2 + g2 = 1942g2 + 36g + 324 - 194 =0

2g2 + 36g + 130 = 0

Now apply the Quadratic Root Formula to solve for g.

1354

1636)2(2

)130)(2(4)36()36(

24

2

2

org

g

g

aacbbg

2g2 + 36g + 130 = 0 s = 18 + (-5) = 13or

s = 18 + (-13) = 5

Now to find s.

Therefore the two numbers are:

-5 and 13,or

5 and -13

Roots Word ProblemsExample 5: (no equation given)Two numbers have a difference of 18. The sum of their squares is 194.What are the numbers?

The new area is:32 + (6 × 8) = 32 + 48 = 80

Therefore, each dimension increased by 2 feet.

A rectangle is 8 feet long and 6 feet wide. If each dimension increases by the same number of feet, the area of the new rectangle formed is 32 square feet larger than the area of the original rectangle.How many feet increased each dimension?

-----8-----

----

6---

--x--

--x-

-

80 = (6 + x)(8 + x)80 = 48 + 14x + x2

0 = x2 + 14x – 32

By factoring and the zero property:0 = (x – 2)(x + 16)x = 2 or x = -16

The dimensions of the new rectangle are6 + x and 8 + x 162

21814

21814

21814

232414

)1(2)32)(1(41414 2

xorx

xorx

x

x

x

By the Quadratic Root Formula:

Mixin’ it up

a) When was the ball at a height of 3.5m?

0 = -5t2 + 20t – 2.5

storst

tort

t

t

t

871.3129.0

1071.1820

1071.1820

1071.1820

1035020

)5(2)5.2)(5(42020 2

This question is looking for t so it gives a specific h. In this case h = 3.5.We will solve this by simplifying and using the quadratic root formulaOne time is on the way up and the other is on the way down.

A ball is thrown and follows the path described by the function h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

3.5 = -5t2 + 20t + 1

The ball is at a height of 3.5m twice: once at t = 0.129sand again at t = 3.871s

b) How high is the ball after 4.0s?

This question is looking for h given the value of t = 4.0

After 4.0 seconds in the air, the ball is 1m off the ground.

h = -5(4)2 + 20(4) + 1h = -80 + 80 + 1h = 1


Mixin’ it up

c) What is the ball’s maximum height?

The question is asking for height so I must know the time. Do I?The word MAXIMUM screams VERTEX!! So I do know the time value…The ball reaches its maximum height 2.0 seconds after being thrownWhat is this height?

Therefore, the ball’s maximum height is 21m, which occurs at 2s.

21020

)5(2)20(

2

t

t

t

abt

2114020

1)2(20)2(5 2

hhh


Mixin’ it up

d) When does the ball hit the ground?

This question is asking for the time so I must know the height… h = 0 of course!Time can’t be negative so this cannot be an answer.

Therefore, the ball h it’s the ground in 4.049 seconds.

0 = -5t2 + 20t + 1

storst

tort

t

t

t

049.4049.0

1049.2020

1049.2020

1049.2020

1042020

)5(2)1)(5(42020 2


Mixin’ it up

e) From what height was the ball thrown?

This question is asking for the height so I must know the time…t = 0 of course!

Therefore, the ball was thrown from a height of 1m.

h = -5(0)2 + 20(0)+ 1h = 1


Mixin’ it up

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Quadratic Word Problems