Quadratic Applications with Solutions. “Word Problems scare me!”

Quadratic Quadratic Applications with Applications with

SolutionsSolutions

Quadratic Quadratic Applications with Applications with

SolutionsSolutions

“Word Problems scare me!”

“Word Problems make me nervous!”

“Word Problems?

I just skip them!”

Don’t Worry! With just 5 easy steps, you can become…

A Word Problem Whiz!

STEPS to Solve Word Problems

1. Read and reread the problem.2. Try to picture the problem.3. Ask what you are looking for?4. Determine which formula you should

use to solve the problem.5. Use the formula to solve and

conclude with an answer in sentence form.

# 26

A submarine travelling in a parabolic arc ascends to the surface. The path of the submarine is described by

where x represents the time in minutes and y represents the depth of the submarine in metres.

,50102 2 xxy

a) How deep is the submarine initially?

What are we trying to find out?

The depth of the submarine at time 0. So we have to substitute 0 for x in the function.

50102 2 xxy50)0(10)0(2 2 y

50yInitially, the sub is 50 m below the surface.

# 26



,50102 2 xxy

b) For how long is the submarine underwater?What are we trying to find out?

The positive root of the quadratic equation which will tell where the sub surfaces on the x “time” line.

050102 2 xx

4

40010010 x

4

51010x

1.8xThe submarine is underwater for approximately 8.1 minutes.

# 26



,50102 2 xxy

c) What is the deepest position from the surface?

The sub’s deepest position was 62.5 m below the surface.

What are we trying to find out?The minimum value of the function that occurs at the vertex.

50102 2 xxy

a

bx

2

Use the Vertex

Formula

4

10x

5.2x

Solve for y

50102 2 xxy50)5.2(10)5.2(2 2 y

5.62y

# 6 The manager of a peach orchard is trying to decide when to

arrange for picking the peaches.If they are picked now, the average yield per tree will be 100 kg, which can be sold for 40¢ per kg. Past experience shows that the yield per tree will increase about 5 kg weekly, while the price will decrease by 1¢ per kg weekly.

a) When should the peaches be picked in order to maximize the revenue, and what will the maximum revenue be?What are we trying to find out?

We are looking for the maximum revenue so we need to determine the vertex, but first we must determine the quadratic function for Revenue.

Revenue = Price X QuantityLet x represent the number of weeks before picking.

40.0100Re, venuetodaypickedIf

)01.40(.)5100(Re, venueweekoneinpickedIf

)01.40(.)5100(Re, xxvenueweeksxinpickedIf

Our quadratic function is

)01.40)(.5100()( xxxR 205.40)( xxxR


arrange for picking the peaches.If they are picked now, the average yield per tree will be 100 kg, which can be sold for 40¢ per kg. Past experience shows that the yield per tree will increase about 5 kg weekly, while the price will decrease by 1¢ per kg weekly.

a) When should the peaches be picked in order to maximize the revenue, and what will the maximum revenue be?What are we trying to find out?

We are looking for the maximum revenue so we need to determine the vertex, but first we must determine the quadratic function for Revenue. 205.40)( xxxR

a

bx

2

Use the

Vertex Formula

1.0

1

x

10x

Solve for y

205.40)( xxxR 2)10(05.1040)( xR

45)( xRThe peaches should be picked in 5 weeks for a maximum revenue of $45.


arrange for picking the peaches.If they are picked now, the average yield per tree will be 100 kg, which can be sold for 40¢ per kg. Past experience shows that the yield per tree will increase about 5 kg weekly, while the price will decrease by 1¢ per kg weekly.b) When will the revenue be zero?

What are we trying to find out?We are looking for the root of the quadratic equation.

205.40)( xxxR Let’s use the Quadratic Formula

a=-.05

b=1

c=40

If the peaches are picked in 40 weeks, the revenue will be $0.

a

acbbx

2

42

)05.(2

)40)(05.(411 2

c

x

)05.(2

91

x 20x

40x

inadmissable

# 48 A chemical power plant is rectangular and has a length of 100

m and a width of 60 m. A safety zone of uniform width surrounds the plant. If the area of the safety zone equals the area of the plant, what is the width of the safety zone?

What are we trying to find out?We are finding the width of the safety zone, which we will represent by x. So we will finding the roots of the equation we determine.

100 m

60 m

Area of Chemical Plant = 60 X 100

= 6000 m2

Area of Safety Zone = Area of Total Space – Area of Chemical Plant

Let x represent the width of the safety zone

= (100 + 2x)(60 + 2x) – (100)(60)

= (6000 + 320x + 4x2 ) – (6000)

= 320x + 4x2

# 48 A chemical power plant is rectangular and has a length of 100

m and a width of 60 m. A safety zone of uniform width surrounds the plant. If the area of the safety zone equals the area of the plant, what is the width of the safety zone?

100 m

60 m

So, Area of the Safety Zone = 320x + 4x2

Solve for x.

60003204 2 xx

But the Area of the Safety Zone must equal the Area of the Chemical Plant

060003204 2 xx01500802 xx

a

acbbx

2

42

2

6000640080 x

2

1240080x

68.9568.15 orx

This solution is inadmissible.

The width of the safety zone is approximately 15.68 m.

Look for clues.

Remember to look for clues in the Word Problem.

Which Critical Point will give you the Answer?

When you’ve got

the Quadratic Function,

determine which formula to use

and WORK IT OUT!

You did it!

You are Word Problem Whizzes!

Quadratic Applications with Solutions. “Word Problems scare me!”

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