Topic 4 Applications of Quadratic Equations Unit 7 Topic 4.
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Topic 4Applications of Quadratic Equations
Unit 7 Topic 4
Recall…
• To determine the x-intercepts (zeros) of a function, you will need to factor the equation into the following form: .
• The x-intercepts (zeros) of the function are and .• The axis of symmetry is the equation of the line that
can be drawn through the centre of the graph. It passes through the midpoint between the two x-intercepts.
• The axis of symmetry can then be used to determine the minimum or maximum value for the vertex.
There are several ways to solve (find the roots/zeros of) a quadratic equation:
• graphically• using the graphing calculator
• algebraically • factoring the standard form, , where a• factoring a difference of squares, • using the quadratic formula:
Example 1Maximizing area using a graphical approach
A rectangular lot is bordered on one side by a stream and on the other three sides by 40 m of fencing. The area of the lot is a maximum.
a) Represent the area of the pen as a quadratic function, where A represents the area and x represents the length of one side of the lot.
b) Sketch the graph of the function. c) What dimensions provide the maximum area for the lot?
Try this on your own first!!!!Try this on your own first!!!!
Example 1: Solution
a) Represent the area of the pen as a quadratic function, where A represents the area and x represents the length of one side of the lot.
stream
The three sides have 40 m of fencing.
Whenever possible, start by drawing a picture and summarize the info.
Start with labeling one of the sides as x.
x
The opposite side is also x.
x
Since all three sides must equal 40, the third side must be 40-2x.
40-2x
Area is given by length time width. Therefore, the quadratic function is:
(40 2 )y x x
Example 1: Solution
b) Sketch the graph of the function.
stream Area
x x
40-2x
length (x)
(40 2 )y x x
area (y)
Example 1: Solution
stream Area
x
maximum
x
Using your calculator, find the maximum.
40-2x
length (x)
(40 2 )y x x
c) What dimensions provide the maximum area for the lot?
area (y)
The maximum area of 200m2 occurs when the x-value is 10m.The two dimensions, then are 10m (x) and 20m (40-2x).
Example 2Solving revenue using a graphical approach
A restaurant is currently selling their burgers for $6. From previous sales, they know that at $6 per burger, they can sell 120. After doing some research, they discover that for each $1 price increase, they will sell 10 less burgers. What should the restaurant charge if they want to make the most profit?a) When working with revenue functions, the
unknown variable represents the number of increases/decreases. Define a variable for this question.
Try this on your own first!!!!
Let x represent the number of increases.
Example 2Solving revenue using a graphical approach
b) Represent the selling price of each bag.
c) Represent the number of burgers sold as a function of the selling price.
d) A revenue function is the number of items sold multiplied by the price of each item. Represent the revenue as a function of the selling price.
Try this on your own first!!!!
Selling price: 6 + 1x
Number of burgers sold: 120 - 10x
Revenue: (120 - 10x) (6 + 1x)
Example 2Solving revenue using a graphical approach
e) What selling price will provide the maximum revenue?f) What is the maximum revenue?
Try this on your own first!!!!
( ) 120 10 6 1R x x x
#of burgers sold selling price
Graph
Revenue Function
Find the maximum
f) The maximum revenue is $810, when there are 3 increases.
120-10(3)=90 burgers solde) 6+1(3)=$9 cost per burger
Example 3Solving revenue using a graphical approach
John sells cotton candy at a carnival and is looking to maximize his profits. He determined that if he decreases the price of the candy by $0.25 per bag, he will sell 25 more bags each day. John currently sells 300 bags at $5.50 per bag. Use a graphical approach to solving this problem. a) Represent the number of bags sold as a function of the
number of price decreasesb) Represent the selling price of each bag as a function of the
number of price decreases.c) Represent the revenue as a function of the selling price. d) What selling price will provide the maximum revenue? What
is the maximum revenue?
Try this on your own first!!!!
Example 3: Solution
a) Represent the number of bags sold as a function of the number of price decreases
The question says that John starts by selling 300 bags of cotton candy at $5.50 per bag. For every decrease, John expects he will sell 25 more bags. For x decreases, this means the number sold will be:
300 25x
Example 3: Solution
b) Represent the selling price of each bag as a function of the number of price decreases.
The question says that John starts by selling 300
bags of cotton candy at $5.50 per bag. For every decrease, John will decrease the price of the cotton candy by $0.25. For x decreases, this means a price of:
5.50 0.25x
Example 3: Solution
c) Represent the revenue as a function of the selling price.
( ) (300 25 )(5.50 0.25 )R x x x
number of bags sold
selling price
Example 3: Solution
d) What selling price will provide the maximum revenue? What is the maximum revenue?
In order to solve for the maximum, we need to graph the function and use the maximum function on the calculator.
# of increases (x)
revenue (y)
The maximum revenue is $1806.25, when there are 5 decreases.300+25(5)=425 bags sold 5.50-0.25(5)=$4.25 cost per bag.
Example 4Solving a quadratic equation algebraically
The path of a water slide can be modelled by the quadratic function , where h is the height above the surface of the water and d is the horizontal distance the slider travels, both measured in feet. Algebraically determine the horizontal distance the slider travels before dropping into the pool.
Try this on your own first!!!!
Algebraically means to solve this using algebra as opposed to graphing. In order to find the horizontal distance the slider travels before dropping into the pool, we need to find the x-intercept.
Example 4: Solution
h (𝑑)=−𝑑2−𝑑+12
Factors of -12:-1 × 12 1 × -12-2 × 6 2 × -6-3 × 4 3 × -4
The factors that have a sum of 1 and a product of -12 are -3 and 4.( ) ( 3)( 4)h d d d
h (𝑑)=−(𝑑¿¿ 2+𝑑−12)¿ To simplify this a bit, we can start by factoring out the negative.
Factor
The x-intercepts are 3 and -4.
The slider enters the water after travelling horizontally 3 feet.
Example 5Solving a quadratic equation algebraically
Sylvia dives from a tower into a pool of water. Her dive can be modelled by the function , where h(t) represents her height above the water, in metres, and t represents time from the start of her dive, in seconds. Algebraically determine how long it takes Sylvia to enter the water, to the nearest tenth of a second?
Try this on your own first!!!!
Example 5: Solution
Start by graphing to get a visual
h (𝑡 )=−4.9 𝑡2+1.5 𝑡+10
The x-intercept here tells the time at which she enters the water (h(t)=0)
Solve algebraicallySince the numbers in the equation are not whole numbers, we cannot factor. Use the quadratic formulaa = -4.9 b = 1.5 c = 10
𝑥=−𝑏±√𝑏2−4𝑎𝑐2𝑎
𝑥=−(1.5)± √( (1.5 )¿¿2−4 (−4.9 ) (10 ))2 (−4.9 )
¿
We are looking for the positive x-intercept (since time cannot be negative). Sylvia enters the water at 1.6 seconds.
Example 6Solving a quadratic equation algebraically
A student council is holding a raffle to raise money. The profit function for the raffle is , where p(c) is the profit and c is the price of each ticket, both in dollars. What ticket price will result in the student council breaking even on the raffle?
Try this on your own first!!!!
Example 6: Solution
𝑝 (𝑐 )=−25𝑐2+500𝑐−350Start by graphing
The student council breaks even as soon as the profit function becomes positive. Therefore, we solve for the x-intercepts using the zero function (2nd Trace 2: Zero).
Solve for the x-intercepts
The 1st x-intercept is at 0.7263815045
Price: $0.73
The 2nd x-intercept is at 19.273618
Price: $19.27
Example 7Writing and solving a quadratic equation algebraically
The area of a ping-pong table is 45 ft2. The length is 4 ft more than the width. Algebraically determine the dimensions of the table?
Try this on your own first!!!!
Example 7: Solution
Area – 45 ft2
Whenever possible, start by drawing a picture and summarize the info.
x
x + 4
Start with labeling the width as x.
The length is 4 more than x: x + 4.
Area is given by length time width. Therefore, the quadratic function is: 45 ( 4)x x
Now we need to solve for x so we can find the dimensions!
Example 7: Solution
45 ( 4)x x 245 4x x 20 4 45x x
Factor using sum and product.Factors of -45:
-1 × 45 1 × -45-3 × 15 3 × -15-5 × 9 5 × -9
Two factors that have a product of -45 and a sum of 4 are 9 and -5.
𝑥=−𝑏±√𝑏2−4𝑎𝑐2𝑎
Since the x-value (representing the width) cannot be negative, we know x = 5The dimensions, then, are 5 ft by 9 ft.
0 ( 5)( 9)x x Solve
9 0
9
x
x
5 0
5
x
x
Need to Know:• To solve quadratic equations, we can use two
approaches:▫ 1) a graphical approach by using the graphing
calculator.▫ 2) an algebraic approach by factoring the standard
form (), factoring a sum of squares ( or by using the quadratic formula ( ).
• A revenue function is the number of items sold multiplied by the price of each item.
• When solving maximization problems, the y -value represents the maximum. You’re ready! Try the
homework from this section.
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