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Mathematical models Section 2.8.

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Presentation on theme: "Mathematical models Section 2.8."— Presentation transcript:

1 Mathematical models Section 2.8

2 Interpreting “y is a function of x”
You will be asked to write “y as a function of x” or “y in terms of x”. This means you want an equation with y isolated on one side and the other side should have only x’s. expl: #32 Water is poured into a container in the shape of a right circular cone with radius 4 feet and height 16 feet. (picture on next slide) Express the volume V of the water in the cone as a function of the height h of the water. Hint: The volume V of a cone of radius r and height h is .

3 relationship between r and h
Now write r in terms of h (solve for r) We now have r in terms of h. Finish #32 in homework.

4 expl: #4 The price p and the quantity x sold of a certain product obey the demand equation a.) Express the revenue R as a function of x. b.) What is the revenue when 100 units are sold? c.) What quantity x maximizes revenue? What is the maximum revenue? d.) What price should the company charge to maximize revenue?

5 1. Define your variables. Part a x = number sold (quantity) p = price per item R(x) = revenue from selling x items at p dollars each Revenue price per item number sold 2. Create a verbal model. You are told that so put that in for p.

6 You want the y value when x is 100. So find
Part b You want the y value when x is 100. So find R(100) = $ Look for where the y value is biggest. This means if we sell 150 units, we’ll make $7,500 and that is the most we can make. Part c Use window [0, 350] x [0, 10,000].

7 So the maximum revenue occurs when we sell 150 items at $50 each.
Part d You know the maximum occurs when x is 150. So you want to know the price when the quantity is 150. You use the fact that you know something about how the price and quantity sold are related. That is, So the maximum revenue occurs when we sell 150 items at $50 each.

8 Plot the point (0, -1) and y = x 2 – 8.
expl: #10 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 grapher to graph d = d(x). e.) For what values of x is d smallest? Plot the point (0, -1) and y = x 2 – 8.

9 Imagine the point P traveling around on the parabola
Imagine the point P traveling around on the parabola. As it moves, the distance from (0, -1) changes. We want to capture this relationship in an equation.

10 We have two points, (0, -1) and P. Since P is on the parabola
y = x 2 – 8, the coordinates of P could be thought of as (x, x 2 – 8). Part a Distance between two points (x1, y1) and (x2, y2)

11 Part b asks for d when x = 0. Plug it in and get 7. Parts b & c
Part c asks for d when x = -1. Plug it in and get approximately 6.08.

12 Parts d & e For parts d and e, we will graph the function For part e, we’re looking for where the y values are smallest (lowest). This graph has two spots, when x = and x = 2.55.

13 These two points are the closest to
Part e These two points are the closest to (0, -1) as you can get on this parabola. A distance of d = 2.60 is the shortest distance from (0, -1) to the parabola.

14 expl: #16 A rectangle is inscribed in a semicircle of radius 2. Let
P = (x, y) be the point in Quadrant 1 that is a vertex of the rectangle and is on the semicircle. a.) Express the area A of the rectangle as a function of x. b.) Express the perimeter P of the rectangle as a function of x. c.) Graph A = A(x). For what value of x is A largest? d.) Graph P = P(x). For what value of x is P largest?

15 Area is length times width.
Perimeter is 2 times length plus 2 times width. The width of the rectangle is The length of the rectangle is

16 Area equals length times width.
Parts a & c Use a window of [-4, 4] x [-5, 5]. We disregard the part of the graph where x is less than 0 since x is a distance. The largest area is gotten when x = 1.41.

17 Perimeter Parts b & d Use a window of [-4, 4] x [-5, 10]. We disregard the part of the graph where x is less than 0 since x is a distance. The largest perimeter is gotten when x = 1.79.

18 2.8 homework: 1, 3, 7, 11, 17, 26 (Find the distance each car is from the intersection at time t hours. Remember distance equals rate times time. Then use the Pythagorean Theorem to find the distance between the cars.), 32 (On slides 2 and 3, we found r in terms of h. To complete the problem, substitute that into the formula they give for the volume of a cone.)


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