1 / 12

Chapter 1 Section 3

Learn how to find the vertex, or intersection point, of two lines using the algebraic method. Examples and step-by-step instructions provided.

garyclarke
Download Presentation

Chapter 1 Section 3

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. Chapter 1 Section 3 Intersection Point of a Pair of Lines Read pages 18 – 21 Look at all the Examples

  2. Note: • Authors present one method of finding point of the intersection of two lines. • There are other algebraic methods that you can use to find the point, so use the algebraic method that you are most comfortable with!

  3. Vertex • A vertex is a point formed by the intersection of two lines. • Plural form: Vertices

  4. Authors Method of Finding a Vertex • Convert each of the two linear equations into standard form • Equate the two expressions that are to the right of each ‘y =‘ • Solve for x • Solve for y, by substituting the value of x that you have found into one of the original equations (it doesn’t matter which one). • The point ( x , y ) is the vertex of the two lines.

  5. Exercise 21 (page 22) • Graph and find the vertices of the feasible set formed by the given system of inequalities: • Solution: Givenx and y intercepts 4 x + y> 8 y> – 4 x + 8 ( 2 , 0 ) , ( 0 , 8 ) x + y> 5 y> – x + 5 ( 5 , 0 ) , ( 0 , 5 ) x + 3 y> 9 y> – 1/3 x + 3 ( 9 , 0 ) , ( 0 , 3 ) x> 0 x> 0 y> 0 y> 0

  6. Exercise 21 Graph x = 0 Not to scale I ( 0 , 8 ) Feasible Set II III IV ( 0 , 5 ) ( 9 , 0 ) y = 0 ( 0 , 3 ) ( 2 , 0 ) ( 5 ,0 ) y = – (1/3) x + 3 y = – x + 5 y = – 4 x + 8

  7. Exercise 21 Graph II x = 0 I Feasible Set y = – 4 x + 8 II y = – x + 5 III y = – (1/3) x + 3 IV y = 0

  8. Vertex I x = 0 y = – 4 x + 8 y = – 4 ( 0 ) + 8 = 8 Vertex I : ( 0 , 8 ) Vertex II y = – 4 x + 8 y = – x + 5 – 4 x + 8 = – x + 5 – 3 x = – 3 x = 1 y = – ( 1 ) + 5 = 4 Vertex II : ( 1 , 4 ) Exercise 21: Find the Vertices

  9. Vertex III y = – x + 5 y = – 1/3 x + 3 – x + 5 = – 1/3 x + 3 – 2/3 x = – 2 x = 3 y = – ( 3 ) + 5 = 2 Vertex III : ( 3 , 2 ) Vertex IV y = – 1/3 x + 3 y = 0 0 = – 1/3 x + 3 x = 9 Vertex IV : ( 9 , 0 ) Exercise 21: Find the Vertices

  10. Supply vs Demand Application • Exercise 25 (page 23) Supply Curve: p = 0.0001 q + 0.05 Demand Curve: p = – 0.001 q + 32.5 Determine the quantity that will be produced and the selling price. Solution: q represent the quantity produced p represent the selling price (in dollars) 0.0001 q + 0.05 = – 0.001 q + 32.5 0.0011 q = 32.45 q = 29,500 which rep p = 0.0001 (29500) + 0.05 = 3 ( $3.00 , 29,500)

  11. Solution continued 0.0001 q + 0.05 = – 0.001 q + 32.5 0.0011 q = 32.45 q = 29,500 29,500 units Now take one of the equation and substitute 29,500 in for q p = 0.0001 q + 0.05 p = 0.0001 (29,500) + 0.05 p = 3 $3.00

  12. Answer • Twenty-nine thousand five hundred units are produced that sell for $3.00 per unit.

More Related