12 multiple-choice questions, progressively harder.
A cinema sells 300300300 tickets at 888 dollars each. Research shows each 111 dollar price increase sells 303030 fewer tickets. What ticket price maximizes revenue?
Solution
Correct answer: A
Let xxx be the number of 111 dollar increases. The price is 8+x8 + x8+x and the tickets sold are 300−30x300 - 30x300−30x, so revenue is R=(8+x)(300−30x)=−30x2+60x+2400R = (8 + x)(300 - 30x) = -30x^2 + 60x + 2400R=(8+x)(300−30x)=−30x2+60x+2400.
x=−602(−30)=1 ⇒ price=8+1=9x = -\frac{60}{2(-30)} = 1 \;\Rightarrow\; \text{price} = 8 + 1 = 9x=−2(−30)60=1⇒price=8+1=9
A price of 999 dollars maximizes revenue.
A ball is thrown upward from a 444 foot ledge with initial speed 484848 feet per second, so h=−16t2+48t+4h = -16t^2 + 48t + 4h=−16t2+48t+4. What is its maximum height?
Correct answer: B
The peak is at the vertex time t=−482(−16)=1.5t = -\tfrac{48}{2(-16)} = 1.5t=−2(−16)48=1.5 seconds. Substitute back for the height.
h=−16(1.5)2+48(1.5)+4=−36+72+4=40h = -16(1.5)^2 + 48(1.5) + 4 = -36 + 72 + 4 = 40h=−16(1.5)2+48(1.5)+4=−36+72+4=40
The maximum height is 404040 feet. The time 1.51.51.5 seconds answers "when", not "how high".
A ball is launched from a 666 foot platform at 808080 feet per second, so h=−16t2+80t+6h = -16t^2 + 80t + 6h=−16t2+80t+6. At what time does it reach its highest point?
The highest point is at the vertex time t=−b2at = -\tfrac{b}{2a}t=−2ab with a=−16a = -16a=−16, b=80b = 80b=80.
t=−802(−16)=−80−32=2.5t = -\frac{80}{2(-16)} = -\frac{80}{-32} = 2.5t=−2(−16)80=−−3280=2.5
The ball peaks at t=2.5t = 2.5t=2.5 seconds. The value 106106106 feet is the maximum height, which answers a different question.
A gym has 400400400 members paying 202020 dollars a month. Each 111 dollar increase in the fee loses 101010 members. What is the maximum monthly revenue?
Correct answer: D
Let xxx be the number of 111 dollar increases. The fee is 20+x20 + x20+x and the membership is 400−10x400 - 10x400−10x, so R=(20+x)(400−10x)=−10x2+200x+8000R = (20 + x)(400 - 10x) = -10x^2 + 200x + 8000R=(20+x)(400−10x)=−10x2+200x+8000.
x=−2002(−10)=10 ⇒ R=(30)(300)=9000x = -\frac{200}{2(-10)} = 10 \;\Rightarrow\; R = (30)(300) = 9000x=−2(−10)200=10⇒R=(30)(300)=9000
The maximum monthly revenue is 900090009000 dollars, at a fee of 303030 dollars with 300300300 members.
Two numbers differ by 666, and their product is as small as possible. What is that least product?
Correct answer: C
Name the numbers xxx and x−6x - 6x−6, so the product is P=x(x−6)=x2−6xP = x(x - 6) = x^2 - 6xP=x(x−6)=x2−6x. Since a=1>0a = 1 > 0a=1>0 this has a minimum at x=−−62=3x = -\tfrac{-6}{2} = 3x=−2−6=3.
P=(3)2−6(3)=9−18=−9P = (3)^2 - 6(3) = 9 - 18 = -9P=(3)2−6(3)=9−18=−9
The numbers are 333 and −3-3−3, and the least product is −9-9−9.
A bus route has 500500500 riders a day at 444 dollars each. Each 111 dollar fare increase loses 505050 riders. What fare maximizes daily revenue?
Let xxx be the number of 111 dollar increases. The fare is 4+x4 + x4+x and the riders are 500−50x500 - 50x500−50x, so R=(4+x)(500−50x)=−50x2+300x+2000R = (4 + x)(500 - 50x) = -50x^2 + 300x + 2000R=(4+x)(500−50x)=−50x2+300x+2000.
x=−3002(−50)=3 ⇒ fare=4+3=7x = -\frac{300}{2(-50)} = 3 \;\Rightarrow\; \text{fare} = 4 + 3 = 7x=−2(−50)300=3⇒fare=4+3=7
A fare of 777 dollars maximizes revenue.
Two numbers xxx and yyy satisfy x+2y=40x + 2y = 40x+2y=40. What is the greatest possible value of their product xyxyxy?
From x+2y=40x + 2y = 40x+2y=40 we get x=40−2yx = 40 - 2yx=40−2y, so the product is P=(40−2y)y=−2y2+40yP = (40 - 2y)y = -2y^2 + 40yP=(40−2y)y=−2y2+40y.
y=−402(−2)=10 ⇒ x=40−20=20,P=20⋅10=200y = -\frac{40}{2(-2)} = 10 \;\Rightarrow\; x = 40 - 20 = 20, \quad P = 20 \cdot 10 = 200y=−2(−2)40=10⇒x=40−20=20,P=20⋅10=200
The greatest product is 200200200, at x=20x = 20x=20 and y=10y = 10y=10.
A plot is fenced against a wall with 848484 metres of fence on three sides. In the largest plot, how long is each side running out from the wall?
Let xxx be each side out from the wall; the area is A=x(84−2x)=−2x2+84xA = x(84 - 2x) = -2x^2 + 84xA=x(84−2x)=−2x2+84x, maximized at the vertex.
x=−842(−2)=844=21x = -\frac{84}{2(-2)} = \frac{84}{4} = 21x=−2(−2)84=484=21
Each side out from the wall is 212121 metres. The value 882882882 is the area, not a side length.
A theater sells 200200200 tickets at 101010 dollars each. Each 111 dollar increase in price sells 101010 fewer tickets. What is the maximum revenue?
Let xxx be the number of 111 dollar increases. Then R=(10+x)(200−10x)=−10x2+100x+2000R = (10 + x)(200 - 10x) = -10x^2 + 100x + 2000R=(10+x)(200−10x)=−10x2+100x+2000, with vertex at x=−1002(−10)=5x = -\tfrac{100}{2(-10)} = 5x=−2(−10)100=5.
R=(10+5)(200−50)=15⋅150=2250R = (10 + 5)(200 - 50) = 15 \cdot 150 = 2250R=(10+5)(200−50)=15⋅150=2250
The maximum revenue is 225022502250 dollars, at a price of 151515 dollars. The 200020002000 is the starting revenue before any increase.
A ball's height in feet is h=−16t2+96t+7h = -16t^2 + 96t + 7h=−16t2+96t+7. At what time is it at its highest point?
The highest point is at the vertex time t=−b2at = -\tfrac{b}{2a}t=−2ab with a=−16a = -16a=−16, b=96b = 96b=96.
t=−962(−16)=−96−32=3t = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3t=−2(−16)96=−−3296=3
The ball is highest at t=3t = 3t=3 seconds. The value 151151151 feet is the maximum height, which answers "how high", not "when".
Of all rectangles with a perimeter of 100100100, the one with the greatest area is a square. What is that greatest area?
With width xxx the length is 50−x50 - x50−x, so A=x(50−x)=50x−x2A = x(50 - x) = 50x - x^2A=x(50−x)=50x−x2. Since a=−1<0a = -1 < 0a=−1<0 it opens downward, so the maximum is at the vertex.
x=−502(−1)=25,A=25(50−25)=25⋅25=625x = -\frac{50}{2(-1)} = 25, \qquad A = 25(50 - 25) = 25 \cdot 25 = 625x=−2(−1)50=25,A=25(50−25)=25⋅25=625
The greatest area is 625625625, from a 252525 by 252525 square.
A rock is thrown upward from a 202020 foot ledge at 323232 feet per second, so h=−16t2+32t+20h = -16t^2 + 32t + 20h=−16t2+32t+20. What is the greatest height it reaches?
The peak is at t=−322(−16)=1t = -\tfrac{32}{2(-16)} = 1t=−2(−16)32=1 second. Substitute back for the height.
h=−16(1)2+32(1)+20=−16+32+20=36h = -16(1)^2 + 32(1) + 20 = -16 + 32 + 20 = 36h=−16(1)2+32(1)+20=−16+32+20=36
The greatest height is 363636 feet, reached at t=1t = 1t=1 second above the ground.
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