Mathematical

1. Mathematical Models Constructing Functions And Optimisation

2. Suppose a farmer has 50 meters of fencing to build a rectangular yard. Express the rectangular area A he can enclose as a function of the length x of a side. Then find the dimensions to make his yard to enclose the maximum area. Total fencing needed would be the perimeter (adding up all sides) Draw a picture x w w x Area of rectangle is length x times width w This is the area as a function of x and w. We want area as a function of x.

3. Suppose a farmer has 50 meters of fencing to build a rectangular yard. Express the rectangular area A he can enclose as a function of the length x of a side. Then find the dimensions to make his yard to enclose the maximum area. Suppose a farmer has 50 meters of fencing to build a rectangular yard. Express the rectangular area A he can enclose as a function of the length x of a side. Then find the dimensions to make his yard to enclose the maximum area. If we solve for w in this equation, we can substitute it in for w in the area equation below. x w w x To find maximum area, we’ll look at the graph.

4. Suppose a farmer has 50 meters of fencing to build a rectangular yard. Express the rectangular area A he can enclose as a function of the length x of a side. Then find the dimensions to make his yard to enclose the maximum area. Suppose a farmer has 50 meters of fencing to build a rectangular yard. Express the rectangular area A he can enclose as a function of the length x of a side. Then find the dimensions to make his yard to enclose the maximum area. x The graph is a parabola that opens down. Put this in a graphing calculator and trace the x where f(x) is at its maximum. Adjust the window until you get a good view. This is on the next screen. w w x

5. (12.5, 156.25) Remember x is the side of the rectangle and f(x) is the area. This would be the x value that would give the maximum area This would be the maximum area. The maximum enclosed area would be 156.25 square meters

6. Another Example • Let P = (x, y) be a point on the graph of y = x2 – 8 • Express the distance d from P to the point (0, -1) as a function of x. • What is d if x = 0? • What is d if x = -1? • Use a graphing utility to graph d = d(x). • For what values of x is d smallest? The first thing to do is draw a picture. We’ll take each part and do it on a slide.

7. -7 -2 -1 1 3 5 7 -6 -5 -4 -3 0 4 6 8 2 • Let P = (x, y) be a point on the graph of y = x2 – 8 • Express the distance d from P to the point (0, -1) as a function of x. • Let P = (x, y) be a point on the graph of y = x2 – 8 • Express the distance d from P to the point (0, -1) as a function of x. This is a parabola vertically translated down 8. Let’s use the distance formula to express the distance from (x, y) to (0, -1) This is a formula for the distance from P to (0, -1) as a function of x and y. We only want it as a function of x so we need another equation relating x and y to solve and substitute for y. (0, -1) (x, y)

8. -7 -2 -1 1 3 5 7 -6 -5 -4 -3 0 4 6 8 2 • Let P = (x, y) be a point on the graph of y = x2 – 8 • Express the distance d from P to the point (0, -1) as a function of x. Since P is a point on the graph of y = x2 – 8, this equation will be true about the relationship between x and y We can then substitute for y in the distance equation above. y = x2 – 8 (0, -1) (x, y)

9. So we have our formula for the distance from P to (0, -1) and we are ready to answer other parts of the question. b) What is d if x = 0? c) What is d if x = -1?

10. d) Use a graphing utility to graph d = d(x). e) For what values of x is d smallest? This is an even function so will also be smallest d at x = - 2.55

11. Two cars are approaching an intersection. One is 2 km south of the intersection and is moving at a constant speed of 30 km per hour. At the same time, the other car is 3 km east of the intersection and is moving at a constant speed of 40 km per hour. The second car is moving along the x axis so its position at any time is changing but can be written as (x, 0) Express the distance d between the cars as a function of time. Using the distance formula, we can find the distance between (x, 0) and (0, y) to find the distance between the two cars Let’s draw a picture putting the cars on a coordinate system letting the origin be the intersection. (3, 0) (0, -2) The first car is moving along the y axis so its position at any time is changing but can be written as (0, y)

12. Two cars are approaching an intersection. One is 2 km south of the intersection and is moving at a constant speed of 30 km per hour. At the same time, the other car is 3 km east of the intersection and is moving at a constant speed of 40 km per hour. Express the distance d between the cars as a function of time. We need to find equations for x and y in terms of t We now have the distance as a function of time The first car is moving along the y axis. Using d = rt we have d = 30t. It started at -2 on the y axis so it’s y axis position is y = -2 + 30t Similarly the second car is moving along the x axis. Using d = rt we have d = 40t. It started at 3 on the x axis but is moving in the negative x direction so it’s x axis position is x = 3 – 40t

13. By looking at the graph of the distance between the two cars, determine if the cars crash at the intersection and if not, find the minimum distance between them. Here is a graph showing t on the x axis and the distance d on the y axis. Looks like the distance gets close to 0 so let’s zoom in and see if it ever is (meaning the cars did crash). They don’t crash and the closest they get is about ¼ km apart.

14. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au