Mathematical Models Constructing Functions

Suppose a farmer has 50 feet of fencing to build a rectangular corral. 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 corral to enclose the maximum area. Draw a picture x w Total fencing needed would be the perimeter (adding up all sides) x w 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.

Suppose a farmer has 50 feet of fencing to build a rectangular corral. 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 corral to enclose the maximum area. x w If we solve for w in this equation, we can substitute it in for w in the area equation below. x w Suppose a farmer has 50 feet of fencing to build a rectangular corral. 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 corral to enclose the maximum area. To find maximum area, we’ll look at the graph.

Suppose a farmer has 50 feet of fencing to build a rectangular corral. 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 corral to enclose the maximum area. x w x w 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.

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. (12.5, ) The maximum enclosed area would be square feet

Another Example Let P = (x, y) be a point on the graph of y = x 2 – 8 a)Express the distance d from P to the point (0, -1) as a function of x. b)What is d if x = 0? c)What is d if x = -1? d)Use a graphing utility to graph d = d(x). e)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.

Let P = (x, y) be a point on the graph of y = x 2 – 8 a)Express the distance d from P to the point (0, -1) as a function of x. This is a parabola vertically shifted down (x, y) (0, -1) Let P = (x, y) be a point on the graph of y = x 2 – 8 a)Express the distance d from P to the point (0, -1) as a function of x. 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.

(x, y) (0, -1) Let P = (x, y) be a point on the graph of y = x 2 – 8 a)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 = x 2 – 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 = x 2 – 8

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

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 =

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

Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 miles per hour. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 miles 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 = t 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

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 mile apart.