Like this presentation? Why not share!

# 0208 ch 2 day 8

## by festivalelmo on Aug 02, 2012

• 79 views

### Views

Total Views
79
Views on SlideShare
79
Embed Views
0

Likes
0
0
0

No embeds

### Categories

Uploaded via SlideShare as Apple Keynote

• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n
• \n

## 0208 ch 2 day 8Presentation Transcript

• 2.6 Modeling with Functions
• 2.6 Modeling with Functions Page 210 We will do these in class: # 2, 6, 10, 14, 18, 22, 30
• 2.6 Modeling with Functions Page 210 We will do these in class: # 2, 6, 10, 14, 18, 22, 30John 14:27 Peace I leave with you; my peace I give toyou. Not as the world gives do I give to you. Let not yourhearts be troubled, neither let them be afraid.
• 2)
• 2) Draw a Picture!
• 2) Draw a Picture! w w+10
• 2) Draw a Picture! A = lw w w+10
• 2) Draw a Picture! A = lw w A = w(w + 10) w+10
• 2) Draw a Picture! A = lw w A = w(w + 10) w+10 or 2 A = w + 10w a quadratic!
• 6)
• 6) Draw a Picture and Label it!!
• 6) Draw a Picture and Label it!! y x
• 6) Draw a Picture and Label it!! y x P = 2x + 2y
• 6) Draw a Picture and Label it!! A = xy y x P = 2x + 2y
• 6) Draw a Picture and Label it!! A = xy y 16 = xy x P = 2x + 2y
• 6) Draw a Picture and Label it!! A = xy y 16 = xy x 16 y= P = 2x + 2y x
• 6) Draw a Picture and Label it!! A = xy y 16 = xy x 16 y= P = 2x + 2y x ⎛ 16 ⎞ P = 2x + 2 ⎜ ⎟ ⎝ x ⎠
• 6) Draw a Picture and Label it!! A = xy y 16 = xy x 16 y= P = 2x + 2y x ⎛ 16 ⎞ P = 2x + 2 ⎜ ⎟ ⎝ x ⎠ 32 P = 2x + x
• 10)
• 10) r
• 10) 2 C = 2π r A = πr r
• 10) 2 C = 2π r A = πr r Find A, in terms of C
• 10) 2 C = 2π r A = πr r Find A, in terms of C 2 C ⎛ C ⎞ r= ∴ A = π ⎜ ⎟ 2π ⎝ 2π ⎠
• 10) 2 C = 2π r A = πr r Find A, in terms of C 2 C ⎛ C ⎞ r= ∴ A = π ⎜ ⎟ 2π ⎝ 2π ⎠ 2 Cπ A= 2 4π
• 10) 2 C = 2π r A = πr r Find A, in terms of C 2 C ⎛ C ⎞ r= ∴ A = π ⎜ ⎟ 2π ⎝ 2π ⎠ 2 Cπ A= 2 4π 2 C A= 4π
• 14)
• 14) No diagram here, but use a variable list
• 14) No diagram here, but use a variable list 1st Number: x 2nd Number: y
• 14) No diagram here, but use a variable list ⎧ x + y = 60 1st Number: x ⎨ 2nd Number: y ⎩ P = xy
• 14) No diagram here, but use a variable list ⎧ x + y = 60 1st Number: x ⎨ 2nd Number: y ⎩ P = xy Find P in terms of x
• 14) No diagram here, but use a variable list ⎧ x + y = 60 1st Number: x ⎨ 2nd Number: y ⎩ P = xy Find P in terms of x y = 60 − x
• 14) No diagram here, but use a variable list ⎧ x + y = 60 1st Number: x ⎨ 2nd Number: y ⎩ P = xy Find P in terms of x y = 60 − x P = x ( 60 − x )
• 14) No diagram here, but use a variable list ⎧ x + y = 60 1st Number: x ⎨ 2nd Number: y ⎩ P = xy Find P in terms of x y = 60 − x P = x ( 60 − x ) or P = 60x − x 2 a quadratic!
• 18)
• 18)
• 18) V=??? so look it up online or in your book ...
• 18) V=??? so look it up online or in your book ... 1 2 V = πr h 3
• 18) V=??? so look it up online or in your book ... 1 2 V = πr h 3 1 2 100 = π r h 3
• 18) V=??? so look it up online or in your book ... 1 2 V = πr h 3 1 2 100 = π r h 3 2 300 = π r h
• 18) V=??? so look it up online or in your book ... 1 2 V = πr h 3 1 2 100 = π r h 3 2 300 = π r h we want h in terms of r
• 18) V=??? so look it up online or in your book ... 1 2 V = πr h 3 1 2 100 = π r h 3 2 300 = π r h we want h in terms of r 300 h= 2 πr
• 22)
• 22) ﬁnd dimensions to maximize area w P=20 l
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w 10 = l + w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w l = 10 − w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w A = w (10 − w ) l = 10 − w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w A = w (10 − w ) l = 10 − w 2 A = 10w − w
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w A = w (10 − w ) l = 10 − w 2 A = 10w − w graph and ﬁnd max (vertex)
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w A = w (10 − w ) l = 10 − w 2 A = 10w − w graph and ﬁnd max (vertex) vertex : ( 5,25 ) (width, area)
• 22) ﬁnd dimensions to maximize area w P=20 P = 2l + 2w l 20 = 2l + 2w A = lw 10 = l + w A = w (10 − w ) l = 10 − w 2 A = 10w − w graph and ﬁnd max (vertex) vertex : ( 5,25 ) (width, area) dimensions are 5’ by 5’
• 30)
• 30) 20 x x 12 12-2x 20-2x
• 30) 20 x x 12 12-2x 20-2xa) V = lwh
• 30) 20 x x 12 12-2x 20-2xa) V = lwh V = (20 − 2x)(12 − 2x)x
• 2030) x x 12 12-2x 20-2xb)
• 2030) x x 12 12-2x 20-2xb) 200 < ( 20 − 2x ) (12 − 2x ) x
• 2030) x x 12 12-2x 20-2xb) 200 < ( 20 − 2x ) (12 − 2x ) x graph y1 = (20 − 2x)(12 − 2x)x y2 = 200
• 2030) x x 12 12-2x 20-2xb) 200 < ( 20 − 2x ) (12 − 2x ) x graph y1 = (20 − 2x)(12 − 2x)x y2 = 200 ﬁnd interval(s) where y1 > y2
• 2030) x x 12 12-2x 20-2xb) 200 < ( 20 − 2x ) (12 − 2x ) x graph y1 = (20 − 2x)(12 − 2x)x y2 = 200 ﬁnd interval(s) where y1 > y2 (1.174, 3.898 ) U (10.928,∞ )
• 30 b) continued. What about the domain?
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ )
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain:
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0 20 − 2x > 0 2x < 20 x < 10
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0 20 − 2x > 0 12 − 2x > 0 2x < 20 2x < 12 x < 10 x<6
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0 20 − 2x > 0 12 − 2x > 0 2x < 20 2x < 12 x < 10 x<6 We must use the most restrictive
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0 20 − 2x > 0 12 − 2x > 0 2x < 20 2x < 12 x < 10 x<6 We must use the most restrictive D: ( 0,6 ) Which eliminates all answers where x>10.928
• 30 b) continued. What about the domain? (1.174, 3.898 ) U (10.928,∞ ) Domain: x>0 20 − 2x > 0 12 − 2x > 0 2x < 20 2x < 12 x < 10 x<6 We must use the most restrictive D: ( 0,6 ) Which eliminates all answers where x>10.928 b) answer is: (1.174, 3.898 )
• 30 c) Maximize Volume of the box
• 30 c) Maximize Volume of the box V = (20 − 2x)(12 − 2x)x which is still in y1
• 30 c) Maximize Volume of the box V = (20 − 2x)(12 − 2x)x which is still in y1 graph & ﬁnd max (vertex) in the domain
• 30 c) Maximize Volume of the box V = (20 − 2x)(12 − 2x)x which is still in y1 graph & ﬁnd max (vertex) in the domain vertex : ( 2.427,262.682 )
• 30 c) Maximize Volume of the box V = (20 − 2x)(12 − 2x)x which is still in y1 graph & ﬁnd max (vertex) in the domain vertex : ( 2.427,262.682 ) which is in the form (x,V)
• 30 c) Maximize Volume of the box V = (20 − 2x)(12 − 2x)x which is still in y1 graph & ﬁnd max (vertex) in the domain vertex : ( 2.427,262.682 ) which is in the form (x,V) max volume is 262.682 cubic inches
• HW # 7“It’s a funny thing about life; if you refuse toaccept anything but the best, you very often get it.” W. Somerset Maugham