4.6 Optimization - Penn Math
11/21/2011
4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
4.1
4.1
1
11/21/2011
Math 103 – Rimmer 4.6 Optimization
Fall 2007 Final Exam
A = x ( y + 60 )
constraint 300 = y + x + ( y + 60 ) + x
A = x (120 − x + 60 )
300 = 2 y + 2 x + 60
A = x (180 − x )
240 − 2 x = 2 y y = 120 − x
A = 180 x − x 2
⇒ y = 30
set
smallest x = 1 largest x = 120
A′ = 180 − 2 x = 0
Dimensions of the pen
⇒ 2 x = 180 ⇒ x = 90
90′ × 90′
x
A( x)
1 90 120
179
Absolute Max.
8100
A ( 90 ) = 90 ⋅ 90
7200
A (120 ) = 120 ⋅ 60
Math 103 – Rimmer 4.6 Optimization
A poster has printed material in the middle and a border around it. The poster must contain 60 sq. cm. of printed material. The left and right margins should be 5 cm., the top and bottom margins should be 3 cm. What should be the dimensions of the printed material in order to minimize the amount of paper used for the poster? constraint
A = ( x + 10 )( y + 6 )
3
poster
x
y
printed material
5
5
material
60 A = ( x + 10 ) + 6 poster x 600 A = 60 + 6 x + + 60 x A = 120 + 6 x + 600 x −1
A′ = 6 − 600 x −2
3 A (1) = 11 ⋅ 66
must
A = xy = 60 printed
A ( 60 ) = 70 ⋅ 7
A (10 ) = 20 ⋅12
Dim. of pr. mat.
10cm. × 6cm.
6−
600 set =0 x2
600 6= 2 x
6 x 2 = 600
x 2 = 100 ⇒ x = 10
Absolute Min.
y=
60 x
smallest x = 1 largest x = 60
x
A( x)
1 10 60
726 240
490
2
11/21/2011
Math 103 – Rimmer
4.6 Optimization A box with a square base and an open top must have a volume of 4000 cubic cm. Find the dimensions of the box that minimize the amount of material being used.
constraint must
A= x 2 + 4 xy base
y
V = x 2 y = 4000
4 sides
y=
4000 A = x2 + 4x 2 x
x
A = x 2 + 16000 x −1
x
A′′ = 2 + 32000 x −3
A′ = 2 x − 16000 x −2
A′′ is always positive
16000 set 2x − =0 x2 16000 = 2x x2
A is always concave up A is continuous for x > 0
The local min. at x = 20 is actually an absolute min.
⇒y=
Dimensions of box
20cm. × 20cm. ×10cm.
x3 = 8000 ⇒ x = 20
General ellipse equation
Distance from ( x, y ) to (1, 0 )
x2 y 2 + =1 a2 b2
d=
( x − 1) + ( y − 0 )
d=
( x − 1)
−a
a
−b
4x + y = 4 ⇒ 2
2
x2 y2 + =1 1 4
4000 4000 = 202 400
⇒ y = 10
Math 103 – Rimmer 4.6 Optimization
Find the points on the ellipse 4 x 2 + y 2 = 4 that are furthest away from the point (1,0) .
b
4000 x2
2
2
2
+ y2
Instead of maximizing the distance d , we maximize its square D = d 2 2 2 The constraint comes from the fact that
D = ( x − 1) + y
the point ( x, y ) is on the ellipse.
2
D = ( x − 1) + 4 − 4 x D′ = 2 ( x − 1) − 8 x
2
4 x2 + y 2 = 4 ⇒ y 2 = 4 − 4 x2
smallest x = −1 largest x = 1
set
2 x − 2 − 8x = 0 −6 x = 2 ⇒ x = y = 4 − 4( 2
1 9
)=
−1 3
x
D ( x)
32 9
−1
4
−1 3
48 9
4 2 ⇒ y=± 3
−1 4 2 −1 −4 2 , and , 3 3 3 3
1 D(
−1 3
)=( ) −4 3
Abs. Max.
0 2
+ 4 − = 169 + 329 = 4 9
48 9
3
11/21/2011
Math 103 – Rimmer 4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
4
11/21/2011
A Norman window has the shape of a rectangle surmounted by a semi-circle. If the perimeter of the window is 30 ft., find the value of x so that the greatest possible amount of light is admitted.
A = xy + π2 ( 2x )
constraint
2
30 = x + y + y + 12 ( 2π
window x 2
π
y
A = xy + 8 x
2
A = x (15 − 12 x − π4 x ) + π8 x 2 1 2
2
π
2
π
A = 15 x − x − 4 x + 8 x
A = 15 x − ( 12 + π8 ) x 2 A = 15 x − ( 4 +8π ) x 2 A′ = 15 − ( 15 = (
4 +π 4
x=
set
4 +π 4
)x 60 4 +π
Math 103 – Rimmer 4.6 Optimization
)x =0
2
x 2
)
30 = x + 2 y + π2 x 30 − x − π2 x = 2 y y = 15 − 12 x − π4 x A′′ = − ( 4+4π ) A′′ is always negative A is always concave down A is continuous for x > 0
The local max. at x =
60 4 +π
is actually an absolute max.
5
4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
4.1
4.1
1
11/21/2011
Math 103 – Rimmer 4.6 Optimization
Fall 2007 Final Exam
A = x ( y + 60 )
constraint 300 = y + x + ( y + 60 ) + x
A = x (120 − x + 60 )
300 = 2 y + 2 x + 60
A = x (180 − x )
240 − 2 x = 2 y y = 120 − x
A = 180 x − x 2
⇒ y = 30
set
smallest x = 1 largest x = 120
A′ = 180 − 2 x = 0
Dimensions of the pen
⇒ 2 x = 180 ⇒ x = 90
90′ × 90′
x
A( x)
1 90 120
179
Absolute Max.
8100
A ( 90 ) = 90 ⋅ 90
7200
A (120 ) = 120 ⋅ 60
Math 103 – Rimmer 4.6 Optimization
A poster has printed material in the middle and a border around it. The poster must contain 60 sq. cm. of printed material. The left and right margins should be 5 cm., the top and bottom margins should be 3 cm. What should be the dimensions of the printed material in order to minimize the amount of paper used for the poster? constraint
A = ( x + 10 )( y + 6 )
3
poster
x
y
printed material
5
5
material
60 A = ( x + 10 ) + 6 poster x 600 A = 60 + 6 x + + 60 x A = 120 + 6 x + 600 x −1
A′ = 6 − 600 x −2
3 A (1) = 11 ⋅ 66
must
A = xy = 60 printed
A ( 60 ) = 70 ⋅ 7
A (10 ) = 20 ⋅12
Dim. of pr. mat.
10cm. × 6cm.
6−
600 set =0 x2
600 6= 2 x
6 x 2 = 600
x 2 = 100 ⇒ x = 10
Absolute Min.
y=
60 x
smallest x = 1 largest x = 60
x
A( x)
1 10 60
726 240
490
2
11/21/2011
Math 103 – Rimmer
4.6 Optimization A box with a square base and an open top must have a volume of 4000 cubic cm. Find the dimensions of the box that minimize the amount of material being used.
constraint must
A= x 2 + 4 xy base
y
V = x 2 y = 4000
4 sides
y=
4000 A = x2 + 4x 2 x
x
A = x 2 + 16000 x −1
x
A′′ = 2 + 32000 x −3
A′ = 2 x − 16000 x −2
A′′ is always positive
16000 set 2x − =0 x2 16000 = 2x x2
A is always concave up A is continuous for x > 0
The local min. at x = 20 is actually an absolute min.
⇒y=
Dimensions of box
20cm. × 20cm. ×10cm.
x3 = 8000 ⇒ x = 20
General ellipse equation
Distance from ( x, y ) to (1, 0 )
x2 y 2 + =1 a2 b2
d=
( x − 1) + ( y − 0 )
d=
( x − 1)
−a
a
−b
4x + y = 4 ⇒ 2
2
x2 y2 + =1 1 4
4000 4000 = 202 400
⇒ y = 10
Math 103 – Rimmer 4.6 Optimization
Find the points on the ellipse 4 x 2 + y 2 = 4 that are furthest away from the point (1,0) .
b
4000 x2
2
2
2
+ y2
Instead of maximizing the distance d , we maximize its square D = d 2 2 2 The constraint comes from the fact that
D = ( x − 1) + y
the point ( x, y ) is on the ellipse.
2
D = ( x − 1) + 4 − 4 x D′ = 2 ( x − 1) − 8 x
2
4 x2 + y 2 = 4 ⇒ y 2 = 4 − 4 x2
smallest x = −1 largest x = 1
set
2 x − 2 − 8x = 0 −6 x = 2 ⇒ x = y = 4 − 4( 2
1 9
)=
−1 3
x
D ( x)
32 9
−1
4
−1 3
48 9
4 2 ⇒ y=± 3
−1 4 2 −1 −4 2 , and , 3 3 3 3
1 D(
−1 3
)=( ) −4 3
Abs. Max.
0 2
+ 4 − = 169 + 329 = 4 9
48 9
3
11/21/2011
Math 103 – Rimmer 4.6 Optimization
Math 103 – Rimmer 4.6 Optimization
4
11/21/2011
A Norman window has the shape of a rectangle surmounted by a semi-circle. If the perimeter of the window is 30 ft., find the value of x so that the greatest possible amount of light is admitted.
A = xy + π2 ( 2x )
constraint
2
30 = x + y + y + 12 ( 2π
window x 2
π
y
A = xy + 8 x
2
A = x (15 − 12 x − π4 x ) + π8 x 2 1 2
2
π
2
π
A = 15 x − x − 4 x + 8 x
A = 15 x − ( 12 + π8 ) x 2 A = 15 x − ( 4 +8π ) x 2 A′ = 15 − ( 15 = (
4 +π 4
x=
set
4 +π 4
)x 60 4 +π
Math 103 – Rimmer 4.6 Optimization
)x =0
2
x 2
)
30 = x + 2 y + π2 x 30 − x − π2 x = 2 y y = 15 − 12 x − π4 x A′′ = − ( 4+4π ) A′′ is always negative A is always concave down A is continuous for x > 0
The local max. at x =
60 4 +π
is actually an absolute max.
5
