1. Mr. Pitt has a peach orchard with 30 trees
Mr. Pitt has a peach orchard with 30 trees. Each tree produces an average of 400 peaches. In order to increase the total peach production of his orchard, Mr. Pitt considers adding more trees. Unfortunately, for each additional tree added, the average production per tree decreases by 10 peaches. Mr. Pitt constructs the following table to analyze the effect of adding trees.
Number of
trees added
Peaches/trees
Total number
of trees
Total peach
production
400 30
30400=12000 1
400-1(10)=390
1+30=31
31390=12090 2
400-2(10)=380
2+30=32
32380=12160 3
400-3(10)=370
3+30=33
33370=12210 4
400-4(10)=360
4+30=34
34360=12240 5
400-5(10)=350
5+30=35
35350=12250 6
400-6(10)=340
6+30=36
36340=12240 7
400-7(10)=330
7+30=37
37330=12210 x
(30 + x)
(400-10x)
Y =(30+x) (400-10x)
y = (30 + x) (400 – 10x)
y = – 300x + 400x –10x2
y = –10x x

2. 1) The XYZ publishing company prints 36 books a day, which are sold for $40 each. The company can print 50 books a day. However, for each additional book printed, the selling price per unit decreases by $1. How many additional books must the company print in order to maximize it daily sales? What is this maximum daily sales figure?
Additional
Books
Selling price
per book
Total number
of books
Daily
Sales
$40 36
36$40=$1440 1
1+36=37
40-1(1)=$39
37$39=$1443 2
2+36=38
40-2(1)=$38
38$38=$1444 3
3+36=39
40-3(1)=$37
39$37=$1443 4
4+36=40
40-4(1)=$36
40$36=$1440 x
x+36
40-x(1)
Y =(x+36)(40-x)
y = (x+36)(40-x)
y = 40x – x – 36x
y = – x2 + 4x

3. 2) A survey conducted during an amateur theatre festival in Sherbrooke indicated that 100 people would attend the festival if the tickets were sold at $4.50 apiece. For each $0.30-decrease in the price of a ticket, 20 more people would attend the festival.

4. # of extra people (20)
Ticket price
Total # of people
Revenue h
0 20
$4.50
100
100$4.50=$450
1 20
$4.50-1($0.30)
100+1(20)=120
120$ 4.20 =$504
2 20
$4.50-2($0.30)
100+2(20)=140
140$ 3.90 =$546
3 20
$4.50-3($0.30)
100+3(20)=160
160$ 3.60 =$576
4 20
$4.50-4($0.30)
100+4(20)=180
180$ 3.30 =$594
5 20
$4.50-5($0.30)
100+5(20)=200
200$ 3.00 =$600
6 20
$4.50-6($0.30)
100+6(20)=220
220$ 2.70 =$594
x 20
100+x(20)
$4.50-x($0.30)
(100+20x)( x)
y = (100+20x)( x)
y = 450 – 30x + 90x – 6x2
y = –6x2 + 60x + 450

5. 3) Mickey Mouse Electronics can sell 300 VCRs at a profit of $60 per unit. Seeking to increase its profits, the company decides to take its VCRs off the market to increase the demand. It calculates that its sales will increase by 50 units for each week that the VCRs are off the market, but that its profits will drop by $5 per unit owing to storage costs. Complete the following table based on this information.
# of weeks
off the market
Profit on
each unit
Number of
units sold
Total
Profit
$60
300
300$60=$18000 1
300+1(50)=350
60-1(5)=$55
350$55=$19250 2
300+2(50)=400
60-2(5)=$50
400$50=$20000 3
300+3(50)=450
60-3(5)=$45
450$45=$20250 4
300+4(50)=500
60-4(5)=$40
500$40=$20000 5
300+5(50)=550
60-5(5)=$35
550$35=$19250 6
300+6(50)=600
60-6(5)=$30
600$30=$18000
What ordered pair corresponds to the maximum profit generated by VCR sales?
How many weeks should the company keep its product off the market in order to earn a maximum profit?
(3,$20250)
3 weeks

6. # of weeks
off the market
Profit on
each unit
Number of
units sold
Total
Profit
$60
300
300$60=$18000 1
60-1(5)=$55
300+1(50)=350
350$55=$19250 2
60-2(5)=$50
300+2(50)=400
400$50=$20000 3
60-3(5)=$45
300+3(50)=450
450$45=$20250 4
60-4(5)=$40
300+4(50)=500
500$40=$20000 5
60-5(5)=$35
300+5(50)=550
550$35=$19250 6
60-6(5)=$30
300+6(50)=600
600$30=$18000 x
300+x(50)
60-x(5)
(300+50x)(60-5x)
y = (300+50x)(60-5x)
y = – 1500x x – 250x2
y = –250x x

7. Additional films reproduced Selling price per reproduction
4) A machine reproduces 10 feature films per day; however, it could reproduce 16 films per day. Each reproduction sells for $300. For each additional copy, the unit price will decrease by $20. Find the corresponding equation by completing the table.
$300 10
10$300=$3000
$300-1(20)=$280
10+1=11
$3080
$300–2(20) = $260
= 12
$3120
300-3(20)=$240
= 13
300-4(20) =$220
10 + 4= 14
Additional films reproduced
Selling price per reproduction
# of films reproduced
Total
Revenue 1 2 3 4 x
300 – 20x
10 + x
(10 +x) (300 – 20x)
y = –20x x

8. 5) A plum orchard with 30 trees yields an average of 500 plums per tree. If more plum trees were added, the yield would decrease by 15 plums per tree. The the equation that describes this situation by completing the table below.
500 30
30500=15000
500-1(15)=485
30+1=31
15035
500-2(15)=470
30+2=32
15040
500-3(15)=455
30+3=33
15015
# of plum trees added
Yield per plum tree
Total number of
plum trees
Total
Yield 1 2 3 x
500-15x
30+x
(30+x)(500-15x)
y = –15x2 + 50x

9. 6) At present, Jupiter Electronics could sell 200 computers at a profit of $300 per unit. The company prefers to wait in order to boost demand for its product. For each week the computers are off the market, The company will sell 40 additional computers, but its profit will decrease by $20 per unit. Find the corresponding equation by completing the following table.
# of weeks
off the market
Profit on
each computer
Number of
computers sold
Total
Profit 1 2 3 4 5 6 x
$300
200
200$300=$60000
300-1(20)=$280
200+1(40)=240
240$280=$67200
300-2(20)=$260
200+2(40)=280
280$260=$72800
300-3(20)=$240
200+3(40)=320
320$240=$76800
300-4(20)=$220
200+4(40)=360
360$220=$79200
300-5(20)=$200
200+5(40)=400
400$200=$80000
300-6(20)=$180
200+6(40)=440
440$180=$79200
300-x(20)
200+x(40)
(200+40x)(300-20x)
y = ?
(200+40x)(300-20x)
= – 4000x x – 800x2
= –800x x

10. The height reached by a ball t seconds after it is thrown up into the air with a speed of 30 m/s is presented by the function, h = -5t2 + 30t. How many seconds will it take for the ball to reach its maximum height? h

11. In this case, we do not determine an increasing and a decreasing intermediate quantity to lead us to the quantity we are ultimately looking for, height. We can use the formula provided which demonstrates the relationship height has to time directly.
45 m
40 m
Time, t
(seconds)
Height, h (meters)
h = -5t2 + 30t
25 m
h = -5(0)2 + 30(0) = = 0 m 1
h= -5(1)2+30(1) = = 25 m 2
h= -5(2)2+30(2) = = 40 m 3
h= -5(3)2+30(3) = = 45 m 4
h= -5(4)2+30(4)= = 40 m 5
h=-5(5)2+30(5)= = 25 m
0 m 6
h=-5(6)2+30(6)= = 0 m

12. What quantity is being referred to as maximum or minimum?
When the situation is presented in this way, there is only a need for two columns. The quantities from these columns are the quantities which illustrate the relationship. These types of relationships have a critical point which is either a maximum or a minimum.
Time, t
(seconds)
Height, h
(meters) 5
25 m 4
40 m 2 3
45 m
0 m 1 6
Which is it in this case?
What quantity is being referred to as maximum or minimum?
At what time does this point occur?

13. Consider the function, h = -5t2 + 20t + 30 representing the height in meters attained by an object t seconds after it is thrown up into the air. Find the time required for this object to reach its maximum height. What is its maximum height?
Time, t
(seconds) 4 2 3 1
Height, h = -5t2 + 20t + 30
(meters)