Max/min Finding Roots

Max/minFinding Roots

You should know the following about quadratic functions:

How to graph them How to find the vertex How to find the x- and y- intercepts How to find the equation from the pattern How to find the equation from the graph How to change from one form to another

There 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

There 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 information to find the equation

These are the harder ones!!

These are the harder ones!!

We have seen that the vertex of a quadratic in general form is given by We have seen that the vertex of a quadratic in general form is given by

a

bf

a

b

2,

2

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.

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.

y-value’s a min when a >0

y-value’s a max when a<0

a. What price would maximize the revenue? a. What price would maximize the revenue?

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

808.0

64

)4.0(2

642

p

p

p

a

bp

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

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

Example 1

A small business’ profits over the last year have been related to the price of the only product. The relationship is R(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.

Example 1


b. What is the maximum revenue possible? b. What is the maximum revenue possible?

The answer to this question is the y-value of the vertex.The answer to this question is the y-value of the vertex.

160)80(

240051202560)80(

24005120)6400(4.0)80(

2400)80(64)80(4.0)80(

802

R

R

R

R

p

The maximum revenue is $160 000.The maximum revenue is $160 000.

Example 1


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

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

2400)0(

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

02

R

R

p

The business would lose $2 400 000.The business would lose $2 400 000.

Example 2: no equation

A farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the area for his livestock. According to regulations the pens need to find these relative dimensions:

Sheep pen

Pig Pen

Kids

What are the dimensions of the largest area?

What are the dimensions of the largest area?



Again the word Maximum means VERTEX. BUT…

Again the word Maximum means VERTEX. BUT…



We know:

The total fencing is 800m. This means

3w+3w+3w+8L+8L+w+2L=800

or 10w + 18L =800This equation has two unknowns!!This equation has two unknowns!!



The question is asking for the maximum area. So let’s get an area equation:

The question is asking for the maximum area. So let’s get an area equation:

Area of a rectangleA = (3w)(8L)A = 24wL

The two equations areThe two equations areA = 24wL 10w + 18L =800

To solve this system of equations, use substitution.

Solve the linear equation (the one without the multiplication of variables) for one of the unknowns.

10w + 18L =800

10w =800 -18L w = 80 -1.8L10w =800 -18L w = 80 -1.8L

Substitute this into the area equation.

A = 24wLA = 24(80-1.8L)L

A = 24L(80-1.8L)A = 1920L-43.2L2

A = 1920L-43.2L2A = 1920L-43.2L2

Remember, we’re looking for the MAX AREA. We’ll find that with the vertex.

Remember, we’re looking for the MAX AREA. We’ll find that with the vertex.

22.224.86

1920

)2.43(2

19202

L

L

L

a

bL

The BEST L value to use is 22.22m

This leads to a w value of

w = 80 -1.8Lw = 80-1.8(22.22)w= 40.00m

w = 80 -1.8Lw = 80-1.8(22.22)w= 40.00m



The dimensions that would be the best are: w = 40.00m and L = 22.22m

The dimensions that would be the best are: w = 40.00m and L = 22.22m

The largest area would be:

A = 24wLA = 24(40.00)(22.22)A = 21331.2m2

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

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

2w + L = 75 A = wL

L= 75-2w

A=w(75-2w)

A = 75w-2w2A = 75w-2w2

75.18

)2(2

752

w

w

a

bw

A = 75(18.75)-2(18.75)2 A =1406.25-703.125A= 703.125

A = 75(18.75)-2(18.75)2 A =1406.25-703.125A= 703.125

We know that we can find the roots of a quadratic function by setting one side equal to zero and

We know that we can find the roots of a quadratic function by setting one side equal to zero and

Factoring (sometimes)Factoring (sometimes)

Completing the square (too long)

Completing the square (too long)

Using the quadratic root formulaUsing the quadratic root formula

This ALWAYS works for a quadratic in general form and is easy to do.

This ALWAYS works for a quadratic in general form and is easy to do.

a

acbbx

2

42

Example 1

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.

a. How long was the duck underwater?a. How long was the duck underwater?

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

)4(20

420 2

xx

xx

x20 40 x

So x = 0 or 4

The duck was underwater for 4 seconds

The duck was underwater for 4 seconds

Example 1

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 from the water and x represents the time in seconds.

b. When was the duck at a depth of 5m?b. When was the duck at a depth of 5m?We can plug in y= -5 and solve for x.

4

244

4

40164

)2(2

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

2

4

2

2

x

x

x

a

acbbx

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.

5420

4252

2

xx

xx

Example 1

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 from the water and x represents the time in seconds.

b. When was the duck at a depth of 5m?b. When was the duck at a depth of 5m?We can check this by finding the minimum value of y.

2

)1(4)1(2

14

4

)2(2

)4(2

2

y

y

x

x

x

a

bx

We conclude that the duck is never 5m below the water.

Example 1

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.

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

We can plug in y= -0.5 and solve for x.

The duck was 0.5m below at t = 0.14s and at t = 1.87s

This will give us the times when the duck is at 0.5 m below.

This will give us the times when the duck is at 0.5 m below.

5.0420

425.02

2

xx

xx

sorx

x

x

x

x

a

acbbx

87.114.0

4

46.344

124

4

4164

)2(2

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

2

4

2

2

Therefore it was below 0.5m for 1.73s


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 unmowed, what is the width of the strip?

40%

The area of the lawn is 8 x 6 =48The area of the lawn is 8 x 6 =48

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

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

The dimensions of this unmowed rectangle are

-----8-2x----- -6-2

x-



40%

So 19.2 = (8-2x)(6-2x)So 19.2 = (8-2x)(6-2x)

-----8-2x----- -6-2

x- We need to solve for

xWe need to solve for x

Make one side equal to 0 and use the quadratic root formula

But first we FOIL it outBut first we FOIL it out



40%

So 19.2 = (8-2x)(6-2x)So 19.2 = (8-2x)(6-2x)

-----8-2x----- -6-2

x-

8.282840

2.19482840

482842.19

41216482.19

2

2

2

2

xx

xx

xx

xxx



40%

0 = 4x2-28x+28.80 = 4x2-28x+28.8

-----8-2x----- -6-2

x-

mormx

x

x

a

acbbx

45.1125.1

8

98.1728

)4(2

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

2

4

2

2

The mowed strip has a width of 1.25m



40%

-----8-2x----- -6-2

x-

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!Which was 40% of the total area!

Roots Word ProblemsRoots Word Problems


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

Let’s define our variables: S = one of the numbers G= the other number

The question indicates two equations relating these two variables.

S – G = 18S – G = 18

S2+ G2 = 194S2+ G2 = 194

Again, we have a substitution situation. Solve the simpler equation for a variable and plug it in to the other equation.

S=18+GS=18+G

(18+G)2 + G2 = 194




(18+G)2 + G2 = 194

We need to solve this equation for G. Use the quadratic root formula

Let’s FOIL and make one side equal to 0.

324+36G+ G2 + G2 = 194

2G2 +36G+324-194 =0

2G2 +36G+130=0 135

4

1636

)2(2

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

2

4

2

2

orG

G

G

a

acbbG

S = 18 + (-5) = 13 or S = 18+(-13)=5




(18+G)2 + G2 = 194

The numbers are either -5 and 13

or -13 and 5

Roots Word Problems: Try Roots Word Problems: Try oneone

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

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

The new are is 32+6x8=32+48=80

The dimensions of the new rectangle are

6+x and 8+x6+x and 8+x

So 80 = (6+x)(8+x)

So 80 = (6+x)(8+x)

80=48+14x+x2

0=x2 +14x-32

x =2 or x = -16

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 and t is the time since the ball was released.


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

3.5 = -5t2 +20t +1 We will solve this by setting one side equal to zero and using the quadratic root formula

0 = -5t2 +20t -2.5

storst

tort

t

t

t

871.3129.0

10

71.1820

10

71.1820

10

71.182010

35020

)5(2

)5.2)(5(42020 2

One time is on the way up and the other is on the way down.

This question is looking for ‘t’ so it gives a specific h. In this case h = 3.5.



a) When was the ball at a height of 3.5m? 0 = -5t2 +20t -2.5

storst

tort

t

t

t

871.3129.0

10

71.1820

10

71.1820

10

71.182010

35020

)5(2

)5.2)(5(42020 2

The ball is at a height of 3.5m at two times: at t= 0.129s and at t = 3.871s



b) How high is the ball after 4.0s?

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

h = -80 +80 +1h = 1

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

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



c) What is the ball’s maximum height? h = -5t2 +20t +1

210

20

)5(2

)20(2

t

t

t

a

btThe question is asking for height so I

must know the time. Do I?

The word MAXIMUM screams VERTEX!!I do know the time value…It’s

a

bt2

The ball reaches its maximum height 2.0 seconds after being thrown

The max height:

21

14020

1)2(20)2(5 2

h

h

h

The maximum height is 21 m



d) When does the ball hit the ground? 0 = -5t2 +20t +1

storst

tort

t

t

t

049.4049.0

10

49.2020

10

49.2020

10

49.202010

42020

)5(2

)1)(5(42020 2

This question is asking for the time so I must know the height.

The height is 0 – hitting the ground!

Time can’t be negative so this cannot be an answer.



e) From what height was the ball thrown?

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

h =1

This question is asking for the height so I must know the time.

The time is 0 – just before it is thrown!

Max/min Finding Roots

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