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 = 12000 – 300x + 400x –10x2 y = –10x2 + 100x + 12000

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 – x2 + 1440 – 36x y = – x2 + 4x + 1440

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.

# 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)(4.5-0.3x) y = (100+20x)(4.50-0.30x) y = 450 – 30x + 90x – 6x2 y = –6x2 + 60x + 450

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

# 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 = 18000 – 1500x + 3000x – 250x2 y = –250x2 + 1500x + 18000

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 10 + 2 = 12 $3120 300-3(20)=$240 10 + 3 = 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 = –20x2 + 100x + 3000

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 + 15000

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) = 60000 – 4000x + 12000x – 800x2 = –800x2 + 8000x + 60000

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

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 + 0 = 0 m 1 h= -5(1)2+30(1) = -5+30 = 25 m 2 h= -5(2)2+30(2) = -20+60 = 40 m 3 h= -5(3)2+30(3) = -45+90 = 45 m 4 h= -5(4)2+30(4)= -80+120 = 40 m 5 h=-5(5)2+30(5)=-125+150= 25 m 0 m 6 h=-5(6)2+30(6)=-180+180= 0 m

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?

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)