1. Topic 4 Applications of Quadratic Equations Unit 7 Topic 4

2. Recall…

4. Example 1 Maximizing 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!!!!

5. 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:

6. Example 1: Solution b) Sketch the graph of the function. stream Area x x 40-2x length (x) area (y)

7. Example 1: Solution stream Area x maximum x Using your calculator, find the maximum. 40-2x length (x) c) What dimensions provide the maximum area for the lot? area (y) The maximum area of 200m 2 occurs when the x-value is 10m. The two dimensions, then are 10m (x) and 20m (40-2x).

8. Example 2 Solving 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.

9. Example 2 Solving 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)

10. Example 2 Solving 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!!!! #of burgers soldselling price Graph Revenue Function Find the maximum f) The maximum revenue is $810, when there are 3 increases. 120-10(3)=90 burgers sold e) 6+1(3)=$9 cost per burger

11. Example 3 Solving 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 decreases b)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!!!!

12. 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:

13. 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:

14. Example 3: Solution c) Represent the revenue as a function of the selling price. number of bags soldselling price

15. 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 sold5.50-0.25(5)=$4.25 cost per bag.

16. Example 4 Solving a quadratic equation algebraically 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.

17. Example 4: Solution Factors of -12: -1 × 121 × -12 -2 × 62 × -6 -3 × 43 × -4 The factors that have a sum of 1 and a product of -12 are -3 and 4. 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.

18. Example 5 Solving a quadratic equation algebraically Try this on your own first!!!!

19. Example 5: Solution Start by graphing to get a visual The x-intercept here tells the time at which she enters the water (h(t)=0) Solve algebraically Since the numbers in the equation are not whole numbers, we cannot factor. Use the quadratic formula a = -4.9b = 1.5c = 10 We are looking for the positive x-intercept (since time cannot be negative). Sylvia enters the water at 1.6 seconds.

20. Example 6 Solving a quadratic equation algebraically Try this on your own first!!!!

21. Example 6: Solution Start 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 (2 nd Trace 2: Zero). Solve for the x-intercepts The 1 st x- intercept is at 0.7263815045 Price: $0.73 The 2 nd x- intercept is at 19.273618 Price: $19.27

22. Example 7 Writing and solving a quadratic equation algebraically The area of a ping-pong table is 45 ft 2. The length is 4 ft more than the width. Algebraically determine the dimensions of the table? Try this on your own first!!!!

23. Example 7: Solution Area – 45 ft 2 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: Now we need to solve for x so we can find the dimensions!

24. Example 7: Solution Factor using sum and product. Factors of -45: -1 × 451 × -45 -3 × 153 × -15 -5 × 95 × -9 Two factors that have a product of -45 and a sum of 4 are 9 and -5. Since the x-value (representing the width) cannot be negative, we know x = 5 The dimensions, then, are 5 ft by 9 ft. Solve

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