Download Chapter 1 Section 3

Chapter 1 Section 3
Intersection Point of a Pair of Lines
Read pages 18 – 21
Look at all the Examples
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!
Vertex
• A vertex is a point formed by the
intersection of two lines.
• Plural form: Vertices
Authors Method of Finding a Vertex
1. Convert each of the two linear equations into
standard form
2. Equate the two expressions that are to the right
of each ‘y =‘
3. Solve for x
4. 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).
5. The point ( x , y ) is the vertex of the two lines.
Exercise 21 (page 22)
• Graph and find the vertices of the feasible set formed by
the given system of inequalities:
• Solution:
Given
x and y intercepts
4x+y>8
y>–4x+8
(2,0) , (0,8)
x+y>5
x+3y>9
x>0
y>0
y>–x+5
y > – 1/3 x + 3
x>0
y>0
(5,0) , (0,5)
(9,0) , (0,3)
Exercise 21 Graph
x=0
Not to scale
I
(0,8)
Feasible Set
II
(0,5)
III
IV
(9,0)
y=0
(0,3)
(2,0)
( 5 ,0 )
y= –4x+8
y = – (1/3) x + 3
y =–x+5
Exercise 21 Graph II
x=0
I
Feasible Set
y=–4x+8
II
y=–x+5
III
y = – (1/3) x + 3
IV
y=0
Exercise 21: Find the Vertices
• Vertex I
x=0
y=–4x+8
y=–4(0)+8=8
Vertex I : ( 0 , 8 )
• Vertex II
y=–4x+8
y=–x+5
–4x+8=–x+5
–3x=–3
x=1
y=–(1)+5=4
Vertex II : ( 1 , 4 )
Exercise 21: Find the Vertices
• Vertex III
y=–x+5
y = – 1/3 x + 3
– x + 5 = – 1/3 x + 3
– 2/3 x = – 2
x=3
• Vertex IV
y = – 1/3 x + 3
y=0
0 = – 1/3 x + 3
x=9
y=–(3)+5=2
Vertex III : ( 3 , 2 )
Vertex IV : ( 9 , 0 )
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
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
Answer
• Twenty-nine thousand five hundred units
are produced that sell for $3.00 per unit.