Mr. Gardner’s MPM2D name
Mr. Gardner's MPM2D
name ______________________________
Applications of Quadratic Functions Word Problems ISU
Part A: Revenue and Numeric Problems
When you solve problems using equations, your solution must have four components: 1. A let statement, table or diagram where you define the variables used to solve the problem. 2. The equations which model the problem. 3. The algebraic solving, showing all steps which you cannot perform in your head 4. A closing statement which answers the question posed in the problem.
One of the most important things you must do is read the question carefully and look for the key words.
Part 1: Revenue Questions
Need to know: Revenue = (number of items sold) x (price of per item)
1. The current price of an amateur theater tickets is $20, and the venue typically sells 500 tickets. A survey found that for each $1 increase in ticket price, 10 fewer tickets are sold. (a) What is the number of $1 increases in price that will maximize the revenue? (b) What price per ticket will maximize the revenue? (c) What is the maximum revenue?
Note: Each question has key word "maximum" so we will need to find the vertex.
(a) and (b): These parts asks for the number of $1 increases, so, let x be the number of $1 increases.
The current revenue before the survey is R = (500) (20) because there are 500 items sold at $20 each.
Now add the survey results into the equation using x as the number of $1 increases.
The price increases for each $1 increments, so the price changes from (20) to (20 + 1x) For each $1 increase, 10 fewer tickets are sold. Tickets sold changes from (500) to (500 - 10x).
The revenue equation based on the survey is R = (500 - 10x) (20 + 1x).
Because the question says "maximize" we need to find the vertex of R = (500 - 10x) (20 + 1x) . Since the equation is already factored (although not fully factored) the fastest way to do this is by using the midpoint of the zeros. Solve each bracket for zero just like any other quadratic equation. The first zero is 500?10=50, and the second is -20. The midpoint of these zeros is (50-20)?2=15. This means that h = 15, and the x-value of the vertex is 15. In short, x = 15.
Conclusion for (a): The theatre should have 15, $1 increases to maximize their revenue.
Conclusion for (b): The new price will be 20 + 1?15, or $35 per ticket.
(c) For part (c) we need to find the k value of the vertex. To do this, substitute x = 15 in the R equation, and calculate R. R=(500-10?15) (20+1?15)=12250 Conclusion for (c): maximum revenue will be $12250.
1
Mr. Gardner's MPM2D
name ______________________________
2. A sticker warehouse sells an average of 6 rolls of stickers per customer at $4 per roll. Statistics
show that for every $0.25 decrease in price, customers will by an additional roll.
(a) According to this model, if the stickers were reduced to $3 per roll, what will be the revenue?
(b) According to this model, at what sticker price will the revenue from stickers be $28.
Set up the Revenue Equation: The revenue equation without the statistics is R = (6) (4). Let x be the number of $0.25 decreases in price. The number sold increases 1 roll for every increment, which changes (6) to (6 + 1x). The price decreases in $0.25 increments which changes the price from (4) to (4 - 0.25x). The revenue equation we will use then, is R = (6 + 1x) (4 - 0.25x)
(a) This question does not ask for the minimum or maximum so we are not finding the vertex. Instead, it asks about the revenue if the rolls cost $3. This represents a $1 reduction in price which is 4 reductions of $0.25. This means that x = 4, so we substitute x = 4 into the equation to revenue equation to calculate the revenue. R = (6 + 1 ? 4) (4 - 0.25 ? 4), R = $3. Conclusion: The revenue will be $30 if the price per roll is $3.
(b) This question also does not ask for a maximum or minimum so we are not finding a vertex. Instead we are given the revenue (R = 28) and asked to find the price. So we sub 28 for R, and solve the equation 28 = (6 + 1x) (4 - 0.25x). First you need to expand this to standard form, R = -0.25x? + 2.5x ? 4, then use the quadratic formula to solve for x. You will get x = 8 or x = 2. This means that there are two prices which will give revenue of $28. The first is 2 decreases of $0.25, and the second a result of 8 decreases of $0.25. Subtract each amount of increments to $4 to get the answers. Conclusion: The price will be $3.50, or it will be $2.00 per roll to have revenue of $28.
3. NTCI team photos sell for $10 each, and the coaches find that they sell on average 30 photographs per team. The coaches do a survey and find out that for each reduction in price of $0.50, an additional two photographs will be sold. (a) At what price will the revenue from the photographs be $150? (b) At what price will the revenue be a maximum?
(a) This question does not say maximum or minimum so we do not find the vertex. You need to set up a revenue equation as we did before, and solve for x when R = 150. Let x be the number of $0.50 reductions in price. The original revenue is R = (30) (10). Two additional photos will be sold for each decrease, so (30) becomes (30 + 2x). The price is reduced in $0.50, so (10) becomes (10 ? 0.50x). The adjusted revenue equation is R = (30 + 2x) (10 - 0.5x) Now, set R = 150, and solve, by expanding and writing the equation in standard form ( -x ? + 5x + 300 = 150) and then factoring or using the quadratic formula. You will get x = 15, or x = -10. There are two answers. The first is x = 15. The price will be 10 - 0.5 ? (15) = $2.50. The second is x = -10. The price will be 10 ? 0.5 ? (-10) = $15 Conclusion: The price will need to be $2.50 or $15.
(b) This question says maximum, which means that we need to find the vertex for R = (30 + 2x) (10 - 0.5x). Since the equation is already factored, we can solve each bracket for zero to find the x-intercept, and find the midpoint. The zeros are -15 and 20. The mid point is 2.5. This means that we need 2.5, decreases of $0.50. 10 - 0.5 ? (2.5) = $8.75. Conclusion: The price that gives the maximum revenue is $8.75
Textbook work: Page 378 #16 you do not need to follow the textbook instructions. You can use my method. Page 391 #18, Page 404 #18, 19
2
Mr. Gardner's MPM2D
Part 2: Numeric Questions
name ______________________________
4. Find two numbers which have a difference of 7, and a product that is a minimum.
Let the larger number be x, and the other be y. Because their difference is 7, x ? y = 7 equation 1 The product (multiplication) of the two numbers is P = (x) (y)
equation 2
The key words are "product" and "minimum" so you must find the vertex of the product equation (equation 2).
To eliminate one variable from equation 2, use equation 1 (we did this in unit 1, systems of equations). equation 1 x ? y = 7. Rearrange it algebraically so it looks like y = x ? 7. Substitute this into equation 2. When you do this, it will look like this: P = x (x ? 7).
This new equation is already fully factored. The fastest way to find the vertex is to solve for the two zeros, and find their midpoint. One of the zeros is 0, the other is 7. Their midpoint is 3.5. This means that the axis of symmetry is at 3.5 (i.e. h = 3.5) and so minimum product occurs when x = 3.5. This answers the question.
To find the second number, substitute x = 3.5 back into equation 1 (x ? y = 7) to find y. y = 3.5 - 7, and you will get y = -3.5.
Conclusion: The two numbers are 3.5 and -3.5.
5. The sum of two numbers is 28. Find the two numbers such that the sum of their squares is a minimum.
Let one number be x and the other y.
Since the sum (add) is 28, x + y = 28 equation 1. The sum (addition) of their squares (exponent 2) is, S = (x)2 + (y)2
equation 2
The key words are "sum of squares, minimum." We need to solve the vertex of equation 2.
Use equation 1 to eliminate y in equation 2. Re-write equation 1 as y = 28 ? x. Equation 2 will now be S = (x)2 + (28 - x)2.
To find the vertex of equation 2, we need to expand and simplify it to standard form, and then find the vertex by completing the square (although you can use any method you wish of the three we have learned).
Equation 2 in expanded form is S = 2x2 - 56x + 784. In vertex form it is S = 2(x - 14)2 + 392. This means that if x = 14 we will get a maximum sum of 392. Substitute x = 14 into equation 1 to solve for y and you will get y = 14 as well.
Conclusion: The two numbers are 14 and 14.
6. The sum of the squares of two consecutive integers is 365. What are the integers?
Let one number be x. Since the two numbers are consecutive, the other number is x + 1. The sum of the squares is (x)2 + (x + 1)2 and it is equal to 365. The equation is (x)2 + (x + 1)2 = 365. If you look at the question, it does not say "maximum", "minimum" so we do not need the vertex. What we need to do, is to solve the equation (x)2 + (x + 1)2 = 365, by factoring or quadratic formula.
3
Mr. Gardner's MPM2D
name ______________________________
.... Turn Over
Expand and simplify the equation into standard form: 2x 2 + 2x ? 364 = 0 Factor or use the quadratic formula to find the two solutions: x = -14, x = 13.
The two solutions to the equation will give us two possible sets of integers which are answers to the question: Conclusion: The first set of integers is -14 and -13, and the second set of integers is 13 and 14.
Textbook work: #17 on page 391. It contains some of the components of question 3, but is a "minimize" question so it also contains some of the components of questions 1 and 2.
Additional Questions
7. The Lazy Boys are planning to have a concert during the Thanksgiving weekend. If the ticket is set to be $75 each, then 1800 tickets will be sold. For each $1 increase in the ticket price, 20 fewer tickets are sold. What should be the price of each ticket to maximum revenue, and what will be the maximum revenue?
8. A sporting goods store sells 90 ski jackets in a season for $200 each. Each $10 decrease in the price would result in five more jackets being sold. Find the number of jackets sold and the selling price to give revenues of $17 600 from sales of ski jackets.
9. The product of two consecutive numbers is 3306. What are the numbers?
10. Two numbers sum to 10. What are the numbers if their product is a maximum?
Hints to textbook questions
Page 378 #16: R=(2000-100x)(10+1x) "maximize" vertex.
Page 391 #18: R=(24800+800x)(1.85-0.05x) maximize vertex
Page 404 #18 : R=10450, so 10450=(4000-40x)(2+0.15x), solve using factoring or formula
Page 404 #19: R=1237.50, so 1237.5=(300-5x)(4+0.10x), solve using factoring or formula
Page 391 #17: minimize vertex
Equation used to substitute and eliminate y: x+y=26 Equation to minimize x2+y2, which becomes x2+(26-x)2=26.
Hints to Additional Questions
7.
R=(1800-20x)(75+1x) Maximum vertex:
8.
Revenue = 17600. 17600=(90+5x)(200-10x), solve the equation by factoring or formula.
9.
(x)(x+1)=3306 expand and simplify to standard form, factor or formula. Factors to (x ? 57) (x + 58).
Answer is 58 and 57, or -58 and -57.
10. x+y=10, re-write as y=10-x. Product is (x ? y), or P=(x)(10-x). To maximize, find the vertex of P=x(1-x).
4
Mr. Gardner's MPM2D
name ______________________________
Applications of Quadratic Functions Word Problems ISU
Part 3: Shapes
Part B - Shapes and Boarder
1. Farmer Joel wants to fence in a rectangular area that has one side bordered by a stream. If he has 80 m of fence, what are the dimensions and the maximum area he can enclose?
y x
stream
Area Equation (to be maximized): A = x y
x Joel only need to fence three sides because the fourth side is the stream. This means that the Fence Equation is: 80 = 2x + y The fence equation will be used to eliminate y in the area equation. Solve for y, and you get y = 80 ? 2x
Substitute y = 80 ? 2 x into A = x y, and you get A = x (80 ? 2x). You can expand this to A = ? 2 x2 + 80x. Because we want the maximum area, put A = ? 2 x2 + 80x in vertex form by completing the square. You get A = -2(x - 20)2 + 800.
Since h = 20, the x dimension in the question is 20 m. Sub x = 20 into y= 80 ? 2x, and you will get y = 40 m. Area is length multiplied by width, so multiply 20 and 40 to get the area, which is 800 m2
Conclusion: The dimensions are 20 m by 40 m, and the area is 800 m2.
2. A handball court is to be surrounded by plastic stripping 30 m long. One side of the court does not need the stripping because it is a brick wall. What must be the dimensions of the court in order to obtain a maximum area? What is the maximum area?
y
x
x
Brick Wall
Area Equation (to be maximized) A = x y
Plastic stripping: 30 = 2x + y. This equation is used to eliminate y in the area equation. Solve this for y and you get y= 30 ? 2x
Substitute y = 30 ? 2 x into A = x y, and you get A = x (30 ? 2x), which expands to A = ? 2 x2 + 30x. Because we want a maximum area, put A = ? 2 x2 + 30x in vertex form by completing the square. and you get A = -2(x ? 7.5)2 + 112.5.
Alternatively, because the equation is already in factored form, you can take the equation A = x (30 ? 2x), find the two zeros, calculate their midpoint, and then sub the midpoint in to solve for the optimal value. The first zero is x = 0. The second is x = 15. The midpoint is 7.5 (this is the h). Sub 7.5 into the equation for x, calculate, and you will get 112.5.
The x dimension is 7.5 m. Sub x = 7.5 into y= 30 ? 2x, and you will get y = 15 m.
Conclusion: The dimensions are 7.5 m by 15 m. The maximum area is 112.5 m2.
5
Mr. Gardner's MPM2D
name ______________________________
3. A rectangular field is to be enclosed by fence and then divided into two by another fence parallel
to one of the sides. If there are 1800 m of fencing available, find the dimensions of the field of
maximum area. What is the area?
x
x
y
y
y
Area Equation (to be maximized) A = 2x y Fencing used to eliminate: 1800 = 4x + 3y, solve for y and y = 600 - 4 x
3
x
x
Substitute y = 600 - 4 x into the area equation: A = 2x 600 - 4x . Because we want the Maximum Area, we
3
3
need to find the vertex.
Because of the fraction, and since it is already factored, the fastest thing to do is use the midpoints of the zeros. One zero is x = 0. Solve the bracket for zero, and get x = 450. The midpoint of 0 and 450 is 225.
The x coordinate of the vertex (or h) is 225. Sub x = 225 in the equation to find the k value of the vertex. The answer is the maximum area 135 000 m2.
Conclusion: The maximum area is 135 000 m2.
4. A rectangular building lot is 8 m longer than it is wide and has an area of 2900m2. Find the dimensions of the lot.
Area = 2900
x
x + 8
Area = 2900. This means that x (x + 8) = 2900 , for this question .
The question does not ask us to find a minimum or maximum. We are not to complete the square. Instead, we are given the dependent variable and must solve the independent variable by factoring or quadratic formula.
x (x + 8) = 2900
x 2 + 8x - 2900 = 0
(x + 58)(x - 50) = 0
x = -58, x = 50
Since x must be a positive number, x = 50 is the only valid answer, and if x = 50, the length (x + 8) must be 58.
Conclusion: The dimensions are 50 m by 58 m.
5. The length of a rectangular field is 2 m greater then three times its width. The area of the field is 1496 m2. What are the dimensions of the field?
Area = 1496
x
3x + 2
x (3x + 2) = 1496
3x 2 + 2x - 1496 = 0
Area = 1496. This means that x (3x + 2) = 1496 , for this question .
The question does not ask us to find a minimum or maximum. We are not to complete the square. Instead, we are given the dependent variable and must solve the independent variable by factoring or quadratic formula.
(x - 22)(3x + 68) = 0
x = 22, x = - 68 3
Since x must be a positive number, x = 22 is the only valid answer, and if x = 22, the length (3x + 2) must be 68.
Conclusion: The dimensions are 22 m by 68 m.
6
Mr. Gardner's MPM2D
name ______________________________
6. A rectangular field is to be enclosed with 920 m of fencing. Prove that the maximum area occurs
when the field is square.
x y
Key word: MAXIMUM. You need to find the vertex. Assume that the field is a rectangle, and follow the same steps as question 1. When you are done, you should see that the length and the width are the same number, which means that the rectangle is in fact, a square.
8. The length of the hypotenuse of a right triangle is 5 cm. Find the lengths of the two sides if one side is 1 cm longer than the other.
5 x
x + 1
This question does not say Maximum or Minimum, so you do not find the vertex. What you need to do is relate the three sides using Pythagorean Theorem, and solve the
equation. x 2 + (x + 1)2 = 52
Expand, simplify, and write in standard form: 2x 2 + 2x - 24 = 0 You can solve this by quadratic formula or factoring.
If we common factor first, then factor, we get 2(x - 3)(x + 4) = 0
This means that x = 3 or x = -4 . Since you cannot have negative dimensions, one side is 3. The other is x + 1, or 4.
Conclusion: The lengths are 3 cm and 4 cm.
Textbook Questions: p. 317 # 17, p. 378 #'s 14, 15 p. 391 #'s 13 ? 16, 20, p. 404 #'s 16, 17, 27
Part 4: Boarders
9. A rectangular picture frame measures 20 cm by 30 cm. A matte is made of uniform width to go inside the frame and make a nice boarder for the picture. The area of the matte is to be equal to the area of the picture. What is the width of the matte?
x 30-2x
This question does not ask for the maximum or minimum so it is not a vertex question, but a quadratic equation question.
Let the uniform width of the matte be x.
The area of the picture frame is (30) (20) = 600
The area of the picture is (30 - 2x )(20 - 2x ).
The area of the matte is the area of the frame subtract the area
30
of the picture: A = (30)(20) - (30 - 2x )(20 - 2x )
x
20-2x
x
x
20
The area of the matte is equal to the area of the picture, so,
(30)(20) - (30 - 2x )(20 - 2x ) = (30 - 2x )(20 - 2x ).
To solve this equation, expand, collect like terms and put it in standard form, then factor or use the quadratic formula to solve for x.
7
Mr. Gardner's MPM2D
name ______________________________
10. A rectangular garden measures 15 m by 24 m. A larger garden is to be made by increasing
each side length by the same amount. The resulting area is to be 1.5 times the original area.
Find the dimensions of the new garden to the nearest tenth of a metre.
x
24
15 x
x 15+2x
24+2x x
This question does not ask for the maximum or minimum so it is not a vertex question, but a quadratic equation question.
Let the length that the garden is increased by, be x.
The original garden's area is: A = (24)(15) .
The garden's area after it is increased is:
A = (24 + 2x )(15 + 2x ).
The new area is 1.5 times the original, so if you multiply the original by 1.5, you will get the area of the new garden.
(24 + 2x )(15 + 2x ) = 1.5(24)(15)
To solve this equation, expand, collect like terms and put it in standard form 4x 2 + 78x - 180 = 0 , then factor or use the quadratic formula to solve for x. x = 2.08478 or x = -21.5848. The negative answer is impossible, since you cannot have a negative dimension. Use x = 2.08 to calculate the new dimensions.
Conclusion: The dimensions are 19.16 m by 28.16 m.
11. An open-topped box is to be made from a rectangular piece of tin measuring 50 cm by 40 cm by cutting square of equal size from each corner. The base area is to be 875 cm2. What is the
volume of the box?
50
x x
50-2x
40
40-2x
x x
50-2x
x x
x x x
40-2x
Let the square cut-out from each corner by x by x. When tin is folded up into a box, its height will be x.
The value of x cannot be 20 cm or larger because if you cut 20 cm from both sides, the 40 cm side will not have any left over tin to fold into a box.
The base of the box will be (50 ? 2x) by (40 ? 2x), and its area is to be
875 cm2: 875 = (50 - 2x )(40 - 2x ).
In standard form, this is 4x 2 - 180x + 1125 = 0 .
Solve this equation using factoring or the quadratic formula to find the value for x. One zero is 7.5, and the other is 37.5. The 37.5 is impossible because if x = 37.5, this would mean you would cut 37.5 from each side (75 in total per side) and you would have negative material left with which to make the box. The only viable answer is x = 7.5.
Now that you know x = 7.5, you can figure out the volume of the box by
using V = x (50 - 2x )(40 - 2x ) .
Conclusion: The volume is 6562.5 cm2.
8
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