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Extra Optimization Problems “Enrichment Problems”

Extra Optimization Problems “Enrichment Problems”

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Page 1: Extra Optimization Problems “Enrichment Problems”

Extra Optimization Problems

“Enrichment Problems”

Page 2: Extra Optimization Problems “Enrichment Problems”

2

19. An isosceles triangle has one vertex at the origin and the other

two at the points where a line parallel to and above the x-axis

intersects the curve f x 12 x . Find the maximum area of the triangle .

A xy 2 3A x 12 x A 12x x

2A ' 12 3x 212 3x

x 2

A " 6x

A " 2 0

max

2f 2 12 2 8

A xy 2 8 16

Page 3: Extra Optimization Problems “Enrichment Problems”

2

20. Find the height of the rectangle with largest area that can be

1inscribed under the graph of y

1 2x

2

2xA 2xy A

1 2x

2

22

2 1 2x 4x 2xA '

1 2x

22

2 22 2

2 1 2x2 4xA '

1 2x 1 2x

__

1

2

1

2

+

1maxat

2

2

1 1y

211 2

2

Page 4: Extra Optimization Problems “Enrichment Problems”

21. Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius 5.

r

0.5h R

h – height of cylinderr – radius of cylinderR – Given radius of sphere

2V r h

3hV 25 h

4

23 hV ' 25

4

3 hV "

2

103

10 3V " 023

22 21

r h 52

Therefore a max500

V3 3

22 h

V 5 h4

23 h25

4

10h

3

Page 5: Extra Optimization Problems “Enrichment Problems”

4. An open-top box with a square bottom and rectangular sides is to have a volume of 256 cubic inches. Find the dimensions that require the minimum amount of material.

2

2

S x 4xy

V x y 256

2

2

256S x 4x

x

2

2

2

1024S x

x1024

S' 2xx

10240 2x

xx 8

y 4

3

2048S" 2 0

x

therefore a min

8 x 8 x 4

yx

x

Page 6: Extra Optimization Problems “Enrichment Problems”

6. A right triangle of hypotenuse 5 is rotated about one of its legs to generate a right circular cone. Find the cone of greatest volume.

x

y 5

2 2 2

2

x y 5

1V x y

3

21

V 25 y y3

3V y25 1

3y

3

2

2

V ' y

0

25

325

3

y5

3

y

225V ' y

3V " 2 y

5 5V " 2 0

3 3

2

2 2 2 5 2 2 5x 25 y x 25 x 5 5

3 33 3

Therefore max

Page 7: Extra Optimization Problems “Enrichment Problems”

8. (calculator required) A poster is to contain 100 square inches of picture surrounded by a 4 inch margin at the top and bottom and a 2 inch margin on each side. Find the overall dimensions that will minimize the total area of the poster.

A PIC xy 100

A POS x 4 y 8

A POS y 8 x 4

1008 x 4

x

400132 8x

x

y 14.142Since f’ changes from neg

to pos, we have a minimum 11.1071 22.142

4

4

2 2x

y

Page 8: Extra Optimization Problems “Enrichment Problems”

210. The graphs of y 25 x , x 0 and y 0 bound a region

in the first quadrant. Find the di

maximum peri

mensions of the rectangle of

that can be inscribed in thismete region.r

P 2x 2y 2P 2x 2 25 x

1/221P' 2 2 25 x 2x

2

2

2xP' 2

25 x

22x 2 25 x

2 2x 25 x 2 25

x2

+_

Therefore max

5

2

5 5

2 25

x2

Page 9: Extra Optimization Problems “Enrichment Problems”

11. (calculator required) Find the dimensions of the rectanglewith maximum area that can be inscribed in a circle of radius 10.

A 4xy2 2x y 100

x 7.071 y 7.071

14.142 14.142

Since f’ changes from pos to neg, we have a

maximum

2A 4x 100 x

Page 10: Extra Optimization Problems “Enrichment Problems”

x14. Find the minimum distance from the origin to the curve y e

2 2D x 0 y 0

1/ 22 2xD x e

CALCULATOR REQUIRED

x 0.426Minimum since f ‘ (x) changes fromneg to pos at –0.426

1/ 22 2 0.426D 0.426 e 0.780

Page 11: Extra Optimization Problems “Enrichment Problems”

2

2

15. (calculator required) Consider f x 12 x for 0 x 2 3. Let

A(t) be the area of the triangle formed by the coordinate axes and the

tangent to the graph of f at the point t,12 t . For what value of t is

A(t) a minimum?

A t1

xy2

f ' x 2x f ' t 2t

2y 12 t 2t x t

2 2, y 12 tI 2f x 0 0 y t 2t t 1

22

, 12 t 2tt 1

xIf y 0 02

xt

t2

2222

t 121 t 12t 12

2 2t 4t

Since A ‘ changes from neg to pos, min area at t = 2

Page 12: Extra Optimization Problems “Enrichment Problems”

2

16. Find the maximum distance measured horizontally between

the graphs of f x x and g x x for 0 x 1.

D y y

1D' 1

2 y

1 11 2 y 1 4y 1 y

42 y

+_

Therefore max

1

41 1 1

D4 4 4