1 / 25

Section 3.3 Quadratic Functions

Section 3.3 Quadratic Functions. Application Problems. Quadratic Functions. Find the dimensions of a rectangular field that can be enclosed with 3000 feet of fence and that has the largest possible area. Quadratic Functions. Step 1: Make a drawing and label sides

necia
Télécharger la présentation

Section 3.3 Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 3.3 Quadratic Functions Application Problems

  2. Quadratic Functions Find the dimensions of a rectangular field that can be enclosed with 3000 feet of fence and that has the largest possible area.

  3. Quadratic Functions • Step 1: Make a drawing and label sides • Step 2: Write some formulas (perimeter and area) x y y x

  4. Quadratic Functions • Step 3: Substitute one equation inside the other. To do this, solve the perimeter equation for x or y and plug into the Area formula. We plug into the Area formula because we are trying to maximize area.

  5. Quadratic Functions • Step 3: We solved for y, now plug y into the formula for area. Substitute into area. Distributive property This is our area function. It’s a quadratic function that opens down because a is negative.

  6. Quadratic Functions • This part you don’t have to do, but just so that you understand what’s going on: If I were to plot my area function, this is what it would look like: A This graph represents how the area changes with different values for x. Remember that x in your area equation represents the length dimension. In this graph, length dimensions are on the x axis, and the corresponding area is on the y axis. x If you want to know what the largest possible area is, then you can easily see from the graph, that the maximum area occurs at the vertex.

  7. Quadratic Functions • Notice that if I look for the x in the graph that corresponds to this vertex, this will be the x that gives me maximum area. This x(max) also represents the length of my enclosure. A X(max) x How do I find x(max)? X(max) is also the x-coordinate of my vertex. I know how to do that.

  8. Quadratic Functions • So far, this is what I have: x= 750ft y y x= 750ft We have just proven that if I make my fencing with 750ft of length, I will get the maximum area possible. How do I find my width size, knowing that I have a total of 3000ft of fencing? You can use common sense or use the formula for perimeter. By the way, using common sense is also using the perimeter formula, but in your head.

  9. Quadratic Functions • Common sense tells me that if I have a total of 3000 ft. of fencing, and I have used 1500 ft, then I have 1500ft left for my two widths. This means that y has to be 1500 divided by 2, y = 750ft. Using the formula for perimeter: x= 750ft y y x= 750ft

  10. Quadratic Functions Find the dimensions of a rectangular field that can be enclosed with 3000 feet of fence and that has the largest possible area. 750ft 750ft. 750ft. 750ft

  11. Quadratic Functions You try: Page 171 #45

  12. Quadratic Functions A vendor can sell 275 souvenirs per day at a price of $2.00 each. The cost to the vendor is $1.50 per souvenir. Each $0.10 profit increase decreases sales by 25 per day. What price should be charged to maximize profit?

  13. Quadratic Functions First of all, let’s understand how to choose my variable: Each 10 cent increase in price will affect, obviously, the price of the souvenir and also the amount of souvenirs I sell. A $0.10 increase in price = 1 (10 cent increase) A $0.20 increase in price = 2 (10 cent increases) A $0.40 increase in price = 4 (10 cent increases) x will represent the amount of ten cent increases

  14. Quadratic Functions Let’s write an equation for profit: P = (amount of profit per item)(amount of items sold) P = (0.50) (275) If a souvenir costs $1.50 and I sell it for $2.00, then my profit is $0.50 per souvenir. It says that I sell 275 souvenirs per day. Careful!!! The amount of items I sell depends on the amount of 10 cent increases I make. Careful!!! The profit per item depends on the amount of 10 cent increases I make. Each $0.10 profit increase decreases sales by 25 per day

  15. Quadratic Functions x will represent the amount of ten cent increases P = (amount of profit per item)(amount of items sold) P = (0.50 + 0.10 x) (275 – 25x) P = (0.50+0.10x)(275-25x) We have our equation for profit and it looks like the x-intercept form of a quadratic function

  16. Quadratic Functions We need this equation to look like the x-intercept form. Factor out the coefficient of x in both parentheses. Multiply the constants together. Now it looks like x-intercept form of a quadratic function!!!!

  17. Quadratic Functions • This part you don’t have to do, but just so that you understand what’s going on: If I were to plot my profit function, this is what it would look like: P This graph represents how the profit changes with different values for x. Remember that x in your profit equation represents the amount of 10 cent increases. In this graph, the amount of 10 cent increases are on the x axis, and the corresponding profit is on the y axis. x If you want to know what the largest possible profit is, then you can easily see from the graph, that the maximum profit occurs at the vertex.

  18. Quadratic Functions • Notice that if I look for the x in the graph that corresponds to this vertex, this will be the x that gives me maximum profit. This x(max) also represents the amount of 10 cent increases. P x X(max) How do I find x(max)? X(max) is also the x-coordinate of my vertex. I know how to do that.

  19. Quadratic Functions • So far, this is what I have: We have just proven that if I make 3 ten cent increases ($0.30), that’s where I’ll have the most profit. The original problem asks: What price should be charged to maximize profit? Price that should be charged is: $2.00 + 3(0.10) = $2.30 And that’s your final answer!!!

  20. Quadratic Functions You try: Page 171 #46

  21. Quadratic Functions A rocket is fired upward from ground level with an initial velocity of 1600 ft/sec. When does the rocket reach its maximum height and how high is it at that time?

  22. Quadratic Functions Use the following equation for the height (in feet) of an object moving along a vertical line after t seconds: where is the initial height (in ft.) and is the initial velocity (ft/sec). The velocity is positive if it is traveling upward, and negative if it is traveling downward.

  23. Quadratic Functions A rocket is fired upward from ground level with an initial velocity of 1600 ft/sec. When does the rocket reach its maximum height and how high is it at that time? Step 1: Fill in the equation with the information you have available.

  24. Quadratic Functions Step 2: find the x coordinate of the vertex to find when the rocket reaches its maximum height. s (50, 40,000) The rocket reaches its maximum height at 50 seconds. The rocket reaches a maximum height of 40,000ft. t

  25. Quadratic Functions You try: Page 172 #51

More Related