2

Module 3

Variation

What this module is about

This module deals with the variation of more than two variables. Both direct and inverse variations may occur in the same problem. Joint variation are quantities that are directly related. But when joint variation is combined with inverse variation, then it is called combined variations.

What you are expected to learn

1. identify relationship involving two or more variables. 2. find the relation and constant of variation 3. apply the concept of proportionality.

How much do you know

A. Using k as the constant of variation, write the equation of variation for each of the following:

1. The area (A) of a parallelogram varies jointly as its base (b) and its altitude (a).

2. The volume (V) of a pyramid varies jointly as its base area (b) and its altitude (a).

3. The area of the circle varies directly as the square of its radius.

4. U varies jointly as the square of m and inversely as n.

5. V varies jointly with l, w and h.

6. The volume (V) of a cube varies directly as the cube of its edge (e).

7. The force (F) needed to push an object along a flat surface varies directly as the weight (w) of the object.

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8. The altitude (h) of a cylinder is inversely proportional to the square of its radius (r).

9. M varies directly as r and inversely as s.

10. Q varies jointly as R and T.

What you will do

Lesson 1

Joint Variation

This lesson deals with another concept of variation, the joint variation.

Some physical relationships, as in area or volume, may involve three or more

variables simultaneously. Consider the area of a rectangle which is obtained from the formula A lw where

l is the length w is the width of the rectangle. The table shows the area in square centimetres for different values of the length and the base.

l 2 4 5 6 6 8 8 10 w 3 3 3 5 7 7 11 13

A 6 12 15 30 42 56 88 130

Observe that A increases as either l or w increase or both. Then it is said that

the area of a rectangle varies jointly as the length and the width.

Consider the area of a triangle, which is obtained from the formula: 1

2A ab

where b is the base and a is the altitude of the triangle. The table shows the area in square centimetres for different values of the base and altitude, both being in centimetres.

b 2 4 4 6 6 8 8 10

a 3 3 5 5 7 7 11 13 A 3 6 10 15 21 28 44 65

Observe that A increases as either b or a increase or both. We say that the area of a triangle varies jointly as the base and the altitude.

4

Examples:

1. Find an equation of variation where a varies jointly as b and c, and 36a when 3b and 4c .

Solution: a kbc

36 (3)(4)k substitute the set of given data to find k

36

12k apply the properties of equality

3k

Therefore, the required equation of variation is: 3a bc

2. z varies jointly as x and y . If 16z when 4x and 6y , find the constant of variation and the equation of the relation.

Solution: z kxy

: 16 (4)(6)k substitute the set of given data to find k

16

24k apply the properties of equality

2

3k

The equation of the variation is: 2

3z xy

3. The area A of a triangle varies jointly as the base b and the altitude a of the

triangle. If 265A cm when 10b cm and 13a cm , find the area of a triangle whose base is 8cm and altitude is 11cm.

Solution: A kab the equation of the relation : 65 (13)(10)k substitute the set of given data to find k

65

130k apply the properties of equality

1

2k

5

The equation of the variation is: 1

2A ab

Therefore, when 11a and 8b , the area of the triangle is

(11)(8)1

2A

244A cm

4. The area A of rectangle varies jointly as the length l and the width w and 2180A cm when 9l cm and 5w cm . Find the area of a rectangle whose length

is 20cmand whose width is 5cm .

Solution : A klw the equation of the relation

: 180 (9)(5)k substitute the set of given data to find k

180

45k apply the properties of equality

4k

Therefore, when 9l cm and 5w cm .

4A lw

4(20)(5)A

2400A cm

5. The volume (V ) of a prism on a square base varies jointly as the height ( h ) and the

square of a side ( s ) of the base of the prism. If the volume is 381cm when a side of

the base is 4cm and the height is 6cm, write the equation of the relation.

Solution: Express the relation as:

2V ks h

281 (4) (6)k substitute the given values for V ,s and h

6

81 (16)(6)k

81

96k reduce to lowest term

27

32k

The equation of variation is 227

32V s h

6. Extending the problem on the previous example, find the volume of the prism if a

side of the base is 7 cm and the height is 12 cm. Solution:

227

32V s h from the previous example

227 (7) (12)

32V substitute the given values for s and h

27

(49)(12)32

V

27(588)

32V

15876

32V

3496.125V cm

7. The volume (V ) of a prism on a square base varies jointly as the

height ( h ) and the square of a side ( s ) of the base of the prism. A. If the volume is 72 cm3 when a side of the base is 3 cm and

the height is 8 cm, write the equation of the relation.

Solution: 2V ks h where k is the constant of variation

272 (3) (8)k substitute the given to find k

7

72 (9)(8)k

72 72k

72

72k

` 1k

The equation of the relation is 2V s h

B. Find the volume when a side of the base is 5 cm and the height is 14 cm.

Solution: 2(5) (14)V substitute the given values for s and h

(25)(14)V

3350V cm

C. By how many percent is the original volume 1V increased if a side is

increased by 10% and the height is 20%. Solution:

Denote the new side by S and the new height by H . As a result; S = 1.1s since the side is increased by 10% and H = 1.2h since the height is

increased by 20%. Then the new volume 2V is:

2

2V S H

2

2 (1.1 ) (1.2 )V s h

2

2 (1.21 )(1.2 )V s h

2

2 1.452V s h

Since 2

1V s h , we may substitute 1V into the results, which gives

2 11.452V V .

The increase in volume is 0.452 of the original volume 1V . To change 0.452

to percentage, multiply it by 100% that will give you:

(0.452)(100%) 45.2%

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The following illustrations are applications of variation in different fields of mathematics like Geometry, Engineering, etc.

Examples:

1. The volume of a right circular cylinder varies jointly as the height and the square of

the radius. The volume of a right circular cylinder, with radius 4 centimetres and height 7 centimetres, is 352 cm3. Find the volume of another cylinder with radius 8 centimetres and height 14 centimetres. r

Solution:

The equation of the relation is 2V khr

From the given set of data: 4r cm

7h cm

352V 3cm

To find k substitute the values above:

2V khr

2

Vk

hr rearranging the equation above

2

352

(7)(4)k

352

(7)(16)k

22

7k simplifying the fraction

To find the volume of a cylinder with r = 8 cm and h = 14 cm:

222 (13)(8)

7V

22(14)(64)

7V

2816V 3cm

h

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2. The horsepower h required to propel a ship varies directly as the cube of its speed s . Find the ratio of the power required at 14 knots to that required at 7 knots. Solution:

The equation of the relation is 3h ks

The ratio of power required at 14 knots to 7 knots is

3

2

3

1

(14)

(7)

h k

h k

3

2

3

1

(14)

(7)

h

h the 'k s cancel out

2

1

2744

343

h

h

2

1

8

1

h

h

3. The pressure P on the bottom of a swimming pool varies directly as the depth d of the water. If the pressure1 is 125 Pa2 when the water is 2 metres deep, find the pressure when it is 4.5 metres deep.

Solution 1: P kd

P

kd

solving for the constant of variation

125

2k since P = 125 when d = 2

62.5k

62.5P d

(62.5)(4.5)P

281.25P Pa

1 Pressure is defined as the force exerted per unit area 2 Pascal (Pa) is the metric unit for pressure

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Solution 2: In this solution, you do not need to find k. The equation P kd

maybe written as P

kd

, meaning that the ratio P

dis a constant. Therefore:

1 2

1 2

P P

d d

2125

2 4.5

P

2

(125)(4.5)

2P

2 281.25P Pa

4. The horsepower required to propel a ship varies directly as the cube of its speed. If the horsepower required for a speed of 15 knots is 10 125, find the horsepower required for a speed of 20 knots.

Solution: let P = required horsepower s = speed, in knots Since P varies directly as s3, you have

3P ks (1)

310,125 (15)k

3

10,125

(15)k

10,125

3, 375k

3k

33(20)P substitute k = 3 and s = 20 in (1)

3(8000)P

24, 000P hp

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5. The weight of a rectangular block of metal varies jointly as its length, width and thickness. If the weight of a 13 by 8 by 6 dm block of aluminum is 18.7 kg, find the weight of a 16 by 10 by 4 dm block of aluminum.

Solution: Let W weight in kilograms

l length in decimeters

w width in decimeters t thickness in decimeters

Since the weight of the metal block varies jointly as its length, width and thickness you have:

W klwt

W

klwt

18.7

(13)(8)(6)k

18.7

576k

Substitute 18.7

576k , 16l , 10w and 4t in the equation W klwt to get

the weight of the desired block:

18.7

(16)(10)(4)576

W

20.8W kg weight of the 16 by 10 by 4 dm block

6. The amount of coal used by a steamship traveling at uniform speed varies jointly as the distance traveled and the square of the speed. If a steamship uses 45 tons of coal traveling 80 km at 15 knots, how many tons will it use if it travels 102 km at 20 knots?

Solution: Let T = number of tons used

s = the distance in miles v = the speed in knots

and then T = k(sv2) (1) hence, when T = 45, s = 80 and v = 15, you have

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45 = k(80)(152) k = ___45___ (80)(225) k = __1__ 400 Substituting this value for k in (1), you have T = __1__ (120)(202) 400 T = 48000 400 T = 120 tons Try this out

A. Translate each statement into mathematical statement. Use k as the constant of variation.

1. P varies jointly as q and r .

2. V varies jointly with l , w and h .

3. The area A of a parallelogram varies jointly as the base b and altitude h .

4. The volume of a cylinder V varies jointly as its height h and the square of the radius r .

5. The heat H produced by an electric lamp varies jointly as the resistance R and the square of the current c .

6. The area A of a parallelogram varies jointly as the base b and altitude a

7. The volume V of a pyramid varies jointly as the base area b and the altitude a .

8. The area A of a triangle varies jointly as one-half the base b and the

altitude h . 9. The appropriate length (s) of a rectangular beam varies jointly as its width (w) and its

depth (d).

10. The area A of a square varies jointly as its diagonals 1d and 2d .

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B. Solve for the value of the constant of variation k, then find the missing value. 1. z varies jointly as x and y and z = 62 when x = 5 and y = 6. a. find z when x = 7 and y = 8 b. find x when z = 72 and y = 4 c. find y when z = 82 and x = 4 2. z varies jointly as x and y . If 3z when 3x and 15y , find z when 6x

and 9y .

3. z varies jointly as the square root of the product x and y . If 3z when 3x

and 12y , find x when 6z and 8y .

4. d varies jointly as o and g. If d = 15, when o = 14 and g = 5, find g when o = 21

and d = 8. 5. q varies jointly as r and s. If q = 2.4, when r = 0.6 and s = 0.8, find q when r = 1.6

and s = .01.

6. d varies jointly as e and l . If d = 2.4, when e = 0.6 and l = 0.8, find d when e = 1.6

and l = .01. 7. x varies jointly as w, y and z. If x = 18, when w = 2, y = 6 and z = 5, find x when

w = 5, y = 12 and z = 3. 8. z varies jointly as x and y. z = 60 when x = 3 and y = 4. Find y when z =80 and x =2 .

9. The weight W of a cylindrical metal varies jointly as its length l and the square of its diameter d

a. If W = 6 kg when l = 6 cm and d = 3 cm, find the equation of variation.

b. Find l when W = 10 kg and d = 2 cm.

c. Find W when d = 6 cm and l = 1.4 cm. 10. The amount of gasoline used by a car varies jointly as the distance traveled and the

square root of the speed. Suppose a car used 25 liters on a 100 kilometer trip at 100 km/hr. About how many liters will it use on a 192 kilometer trip at 64 km/hr?

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C. What did the pig say when the man grabbed him by the tail?

Directions: Answer the questions below then transfer the letter associated to each question to the box which contains the correct answer.

I If z varies jointly as x and y, and z = 24, when x = 2 and y = 4, find z when x = 2 and y = 5

S Z varies jointly as x and y and z = 60 when x = 3 and y = 4. Find y when z = 80 and x = 2.

N If z varies jointly as x and y and z = 12, when x = 2 and y = 4, find the constant of variation.

E If w varies jointly as x and y and w = 36 when x = 3 and y = 4, find the constant of variation.

S If z varies jointly as x and y and z = 24, when x = 3 and y = 4, find z when x = 3 and y = 2.

T If A varies jointly as l and w and A is 36 when l = 9 and w = 2, find A if l = 6 and w = 4.

H If a varies jointly as c and d, and a = 20, when c = 2 and d = 4. Find d when a = 25 and c = 8.

O If w varies jointly as x and y2 and w = 24 when x = 2 and y = 3, find the value of w when x = 9 and y = 4.

E If z varies jointly as x and the square of y and z = 20, when x = 4 and y = 2. Find z when x = 2 and y = 4.

H If varies jointly as w2 and l and A = 48 when w = 3 and l = 4, find the value of A when w = 9 and l = 15.

T If z varies jointly as x and the square of y and z = 40, when x = 5 and y = 4. Find z when x = 4 and y = 5.

M x varies jointly as and y and z = 48, when x = 4 and y = 3, find the constant of variation.

F If y varies directly as x and if y = 15 when x = 5, find the value of y if x = 7.

I p varies jointly as r and s and p = 32 when r = 3 and s = 2. Find the constant of variation.

E If w varies jointly as x and y and if w = 15 when x = 2 and y = 3, find the value of w if x = 3 and y = 4.

D If y varies directly as x and y = 6 when x = 8, what is the value of y when x = 24?

5

4

30 16

3

8

12

50

4

3

3

5

2

3

2

18

48

192

21

4

40

40

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Lesson 2

Combined variation Combined variation is another physical relationship among variables. This is the

kind of variation that involves both the direct and inverse variations. This relationship among variables will be well illustrated in the following examples.

Examples: A. The following are mathematical statements that show combined variations.

1. I

kPrt

2. E

kIR

3. c

kar

4. Pv

kt

5.

2abk

c

B. Translate each statement into a mathematical statement. Use k as the constant of

variation.

1. T varies directly as a and inversely as b. ak

Tb

2. Y varies directly as x and inversely as the square of z. 2

kxY

z

3. P varies directly as the square of x and inversely as s.

2kxP

s

4. The time t required to travel is directly proportional to the

temperature T and inversely proportional to the pressure P. kT

tP

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5. The pressure P of a gas varies directly as its temperature t and inversely as its volume V. The following examples are combined variation where some terms are unknown and

can be obtained by the available information.

C. If z varies directly as x and inversely as y, and z = 9 when x = 6 and y = 2, find z when x = 8 and y = 12.

Solution:

The equation is kx

zy

Substituting the given values:

6

92

k

9

3k

3k

(3)(8)

12z

2z

D. x varies directly as y and inversely as z . If 15x when 20y and 40z , find x

when 12y and 20z .

Solution: The equation is ky

xz

Substituting the given values to find k where 15x when 20y and 40z ,

20

1540

k

(15)(40)

20k

30k

ktP

V

17

To find x when 12y and 20z

Using the equation ky

xz

(30)(12)

20x

18x

E. t varies directly as m and inversely as the square of n . if 16t when 8m and 2n , find t when 13m and 3n .

Solution:

The equation of the variation: 2

kmt

n

To find k , where 16t , 8m and 2n , substitute the given values

2

(8)16

(2)

k

216(2)

8k

(16)(4)

8k

64

8k

8k

To find t when 13m and 3n

2

(8)(13)

(3)t

104

9t or

5

119

t

18

F.. r varies jointly as s and t and inversely as u . If 3

28r when 10s , 3t and

56u , find r when 6s , 7t and 84u . Solution:

The equation of the variation: kst

ru

Substitute the given values to find k :

3 (10)(3)

28 56

k

(3)(56)

(28)(10)(3)k

2

10k

1

5k

To find r when 6s , 7t and 84u .

kst

ru

1(6)(7)

5

84r

42 1

5 84r

1

10r

G. Given: w varies directly as the product of x and y and inversely as the square of

z . If 9w when 6x , 27y and 3z , find w when 4x , 7y and 2z .

Solution:

The equation: 2

kxyw

z

Substituting the first given set of values to the equation, where 9w , 6x , 27y and 3z

19

2

(6)(27)9

(3)

k

(162)

99

k

81 162k

81

162k or

1

2k

Find the value of w when 1

2k and use the second set of values when 4x ,

7y and 2z , you have

2

kxyw

z

2

1(4)(7)

2

2w

(2)(7)

4w

7

2w or

3.5w

H. The current I varies directly as the electromotive force E and inversely as the

resistance R . If in a system a current of 20 A flows through a resistance of 20 with an electromotive force of 100 V, find the current that 150 V will send through the system.

Solution: Let I = the current in A (ampere) E = electromotive force in V (volts)

R = ( ohms)

The equation: kE

IR

Substitute the first set of given data:

20

I = 20 A E = 100 V

R = 20 By substitution, find k:

100

2020

k

(20)(20)

100k

400

100k

4k

To find how much (I) current that 150 V will send through the system

(4)(150)

20I

30I

Notice, the system offers a resistance of 20 .

Try this out A. Using k as the constant of variation, write the equation of variation for each of the following.

1. W varies jointly as the square of a and c and inversely as b. 2. The electrical resistance (R) of a wire varies directly as its length ( l ) and inversely

as the square of its diameter (d). 3. The acceleration A of a moving object varies directly as the distance d it travels

and varies inversely as the square of the time t it travels. 4. The heat H produced by an electric lamp varies jointly as the resistance R and

the square of the current C. 5. The kinetic energy E of a moving object varies jointly as the mass m of the object

and the square of the velocity v.

B. Solve the following 1. If r varies directly as s and inversely as the square of u, then r = 2 when

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s = 18 and u = 2. Find:

a. r when u = 3 and s = 27. b. s when u = 2 and r = 4 c. u when r = 1 and s = 36

2. p varies directly as q and the square of r and inversely as s.

a. write the equation of the relation b. find k when p = 40, q = 5, r = 4 and s = 6 c. find p when q = 8, r = 6 and s = 9 d. find s when p = 10, q = 5 and r = 2.

3. w varies directly as xy and inversely as v2 and w = 1200 when x = 4, y = 9 and v =

6. Find w when x = 3, y = 12 and v = 9.

4. Suppose p varies directly as 2b and inversely as 3s . If 3

4p when 6b and

2s , find b when 6p and 4s .

Let s summarize

Definition:

Joint Variation: The statement a varies jointly as b and c means

a kbc , or a

kbc

, where k is the constant of variation.

Combined Variation:

The statement t varies directly as x and inversely as y

means kx

ty

, or ty

kx

, where k is the constant of variation.

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What have you learned

A. The following are formulas and equations which are frequently used in mathematics and in science. State whether the relationship is considered direct, inverse, joint or combined variation.

____________________1. C = 2 ____________________2. A = lw ____________________3. D = rt ____________________4. I = prt ____________________5. V = lwh

____________________6. A = r2

____________________7. V = r2 ____________________8. E = mc2 ____________________9. F = ma

___________________10. V = 2r T

_____________________11. K = 21

2mv

_____________________12. P = F A

_____________________13. W = Fd

_____________________14. V = IR

_____________________15. Q = mct

B. For each given equation with k as the constant of variation and solve for the unknown value. Choose the letter of the correct value of the unknown.

1. w varies inversely as z and w is 4 when z is 6. what is z when w is 14? a. 12/7 b. 11/7 c. 10/7 d. none of these 2. x varies jointly as y and z. x is 4 when y is 3 and z is 2. What is z if x is 8 and y is

10? a. 5/6 b. 6/5 c. 6 d. 5 3. m varies directly as n but inversely as p. What is n if m is 16 and p is 18? a. 284 b. 286 c. 288 d. 290 4. p varies inversely as q and r, and p = 2/3 when q is 4 and r is 14. What is q when p

is 6 and r is 10? a. 45/28 b. 40/25 c. 42/25 d. 28/45

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5. F varies directly as g and inversely as the square root of the product of I and h, and F = 5 when g = 7.5, I = 2 and h = 18. What is F when g = 4, I = and h = 16? a. 8 b. 16 c. 18 d. none of these 6. W varies directly as u2 and v, and W is 75 when u is 5 and v is 9. What is v when W

is 150 and u is 10? a. 4 b. 4.5 c. 5 d. none of these 7. S varies directly as t and inversely as u2, and S is 9 when t is 4 and u is 12. What is t

when u is 8 and S is 16? a. 3.1 b. 1.16 c. 3.16 d. none of these 8. y varies directly as w2 and inversely as the cube of x. What is y when w is 3 and x is 2? a. 8/9 b. 9 c. 8 d. 9/8 9. P varies as the product r and inversely as the square of t. What is t when P = 2, r = 33?

a. 33

2 b. 33/2 c. 11/2 d. none of these

10. y varies directly as x, and y = 6 when x = . What is x if y = 12? a. 3 b. 1 c. 2 d. 0

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Answer Key

How much do you know A. 1. A = kba or A = kab

2. V = kba or V = kab

3. A = kr2

4. 2km

Un

4. V = klwh

V = ke3

7. F = kw

8. h = 2

k

r

9. kr

ms

10. Q = kRT

Try this out Lesson 1 A. 1. P = kqr

2. V = klwh

3. A = kbh

4. V = khr2

5. H = 2krc

6. A = kab

7. V = kab

8. A = 1

2bh

9. s = kwd

10. A = kd1d2

B. 1. Z kxy , 31

15k

a. 115.7 b. 8.71 c. 9.92

2. 1

15k , 3.6z

3. z k xy , 1

2k , 18x

4. 3

14k , 1.8g

5. 50k , 0.8q

6. 5k , 0.08d

7. x kwy , 3

10k or 0.3, 54x

8. z kxy , 5k , 8y

9. a. 1

9k , 2

1

9W ld

b. 22.5l cm c. 5.6 kg

10. 0.025k , 38.4 liters

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C. What did the pig say when the man grabbed him by the tail?

Lesson 2 A.

1. 2ka c

Wb

2. 2

klR

d

3. A = kd

t2

4. 2H kRC

5. E = kmv2

B.

1. 2

ksr

u ,

4

9k

a. r = 4/3 b. s = 36

d. u = 4

2. a. 2kqr

Ps

3. k = 1200

1

1333

w

4. 1

6k

b = 48

b. k = 3 c. P = 96 d. s = 6

H

5/4

I

30 S

16/3

I

8

S

12

T

50

H

4/3

E

3

E

5/2

N

3/2

D

18

T

48

O

192

F

21

M

4

E

40

26

What have you learned A. B. 1. direct 1. a 2. joint 2. b 3. joint 3.c 4. joint 4. d 5. joint 5. a 6. direct 6. b 7. direct 7. c 8. joint 8. d 9. joint 9. a 10. combined 10. b 11. joint 12. combined 13. joint 14. joint 15. joint Example 5. If the volume of a mass of gas at given temperature is 56 in3 when

the pressure is 18 lb/in2, use Boyles law to find the volume when the pressure is 16 lb/in2 Solution: Boyles law states that V = k/P or PV = k, meaning that PV is a constant.

a) without finding k, you may write P1V1 = P2V2 where P1 = 18 lb/in2, V1 =, 56 in3, P2= 16lb/in2, V2 = ? Substituting (56)(18) = (16) V2 V2 = 56(18) 26 V2 = 63 in 3

Example 8. The load which can be safely put on a beam with a rectangular cross section that is supported at each end varies jointly as the product of the width and the square of the depth and inversely as the length of the beam between supports. If the safe load of a beam 3 in wide and 6 in deep with supports 8 ft apart is 2700 lb, find the safe load of a beam of the same material that is 4 in wide and 10 deep with supports 12 ft apart.

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Solution: let w = width of beam, in inches d = depth of beam, in inches l = length between supports, in feet L = safe load, in pounds then L = kwd2 l According to the first set of data ,when w = 3, d= 6 and l = 8, then L = 2700, therefore 2700 = k (3) (62) 8 k = 8(2700) = __8(2700)__ 3(62) 108 k = 200 Consequently, if w = 4, d = 10, l= 12 and k = 200, you have L = 200(4)(102) 12 L = 6666 2/3