Problems in Advanced Algebra

7/25/2019 Problems in Advanced Algebra

1/26

Problem 1

A company manufactures two products, A and B , and it requires three different machines to

process each product. Product A requires 10 hr of time on machineM

1 , 6 hr of time on

machine M2 , and 12 hr of time on machine M3 . Product B requires 10 hr of time on

M1 , 12 hr onM

2 , and 4 hr onM

3 . If the profit of each unit of product A is $ 400

and the profit of each unit of product B is $ 720

, how many units of each product should be

produced in each two-week period if there are 240 hr of time available on each machine and the

company wishes to maximie the profit!

"olution#

It helps to tabulate all the iven information as follows,

Product A Product % &otal 'umber of (ours

)achine1 10

hours 10

hours 240

hours

)achine2 6

hours 12

hours 240

hours

)achine3 12

hours 4

hours 240

hours

)inimum Production 1unit

1unit

Profit $ 400 $ 720

*et x be the number of units of product A produced, y be the number of units of product

B produced, and z be the profit. %ased from the tabulated data above, we have the system of

inequalities,

{10x+10y 2406x+12y 24012x+4y 240

x 1

y 1

+e must find the number of units of each product that maximies the profit, z=400x+720y , within

the reion defined by the system of inequalities above. +e sketch the reion defined by the system of

inequalities above like so,

&he coordinates of the vertices are obtained by solvin the system of equation of two lines intersectin

at a vertex. &he reion has five vertices with coordinates (1,1) , (1, 392) , (8,16 ) , (18,6 ) , and


2/26

( 593 , 1) . %y &heorem , the values of x and y that

maximie or minimie z=ax+by occurs at the vertices of

the reion. +ith this in mind, we evaluate z on these

vertices as follows#

at (1,1 ) , z=400 (1 )+720 (1 )=1120

at(1,

39

2)

,z=400 (1 )+720( 392)=14440

at (8,16 ) , z=400 (8 )+720 (16 )=14720

at (18,6 ) , z=400 (18 )+720 (6)=11520

at(

59

3, 1)

,z=400 (593)+720 (1 )=

25760

38586.67

&herefore, the maximum profit is achieved by producin

8 units of product A and 16 units of product %.

Problem 2

A distributor of video recorders has two warehouses that supply three different retailers. &o deliver a

recorder to retailerR

1 costs $ 27 from warehouseW

1 and $ 36 from warehouseW

2 . It

costs $ 9 to deliver a recorder from W1 to retailer R2 and $ 6 to deliver one from W2 to

R2 . or retailer

R3 , it costs $ 15 if the recorder comes from

W1 and $ 30 if it comes from

W2 . "uppose that

R1 orders three recorders,

R2 orders four recorders, and two recorders are

ordered fromR

3 . If the distributor has five recorders in stock inW

1 and four inW

2 , how many

recorders should be shipped from each warehouse to each retailer in order for the distributor to minimiethe delivery costs!

"olution#

It helps to tabulate all the iven information as follows,


3/26

Problem 3

A class consists of 12 boys and 9 irls. In how many ways could they choose a president,

a secretary and a treasurer if the secretary is a boy and the treasurer is a irl! Provide possible

approaches, determine if the approaches have different answers and ustify the result.

/. &he first approach is to choose the secretary or the treasurer first before the president. It is done

this way because the position for secretary is exclusive for a boy and for treasurer is exclusive

for a irl. or example, since we have 12 boys, we have twelve possible ways of which one

of them will be placed for the position of secretary. "imilarly, we have 9 irls so we have

nine possible ways of which one of them will be placed for the position of treasurer. 'otice that

if I have chosen a boy for the position of secretary, there are still nine choices from which of the

nine irls will be placed for the position of treasurer because the respective positions are enderspecific. &he position for the president is not ender specific0 it does not matter whether I placed

a boy or a irl. "ince I have chosen two people in place of the secretary and treasurer position

and in total there were12+9=21 people, I only have

212=19 people to choose for the

position of president. %y the undamental Principle of 1ountin, the number of ways I could

place people in their respective positions from the class is iven by,


4/26

12919=2052 ways.

. +e could have placed a people for the position of president first before that of secretary and

treasurer. +e must take note what the ender of the chosen president is because it matters for the

next positions due to their ender exclusivity. It is not possible to tell the ender of the chosen

president and it may result to difficulty of determinin how many people are left to choose from

for the next positions, so we take cases separately0 if the president is a boy or a irl.

If we are to choose a president that is a boy, there are 12 possible choices. 2pon choosin, it

follows that there are only121=11 boys to choose from for the position of secretary. "till,

there are9

irls to choose from for the position of treasurer. Applyin the undamental

1ountin Principle, we have,

12119=1188 ways.

If we are to choose a president that is a irl, there are 3 possible choices. 2pon choosin, it

follows that there are only91=8 irls to choose from for the position of secretary. "till,

there are 12 boys to choose from for the position of secretary. Applyin the undamental

Principle of 1ountin, we have,

9 8 12=864 ways.

%y the rule of sum, the total number of ways we could place people on these positions takin the

constraints into consideration is iven by1188+864=2052 ways. "ince we obtained the

same result, this tells us that the second approach ouht to be the same with the first one.

Exam 1

/. "olve the followin inequalities#

4a5 6x2+x>12

4b5 |32x|5

4c5x

2 (x+3 )( 4x3 ) (x2 )

0

4d5|2x+1x3| 4


5/26

"olution#

4a5 irst, we need to make sure that at least one side if the inequality is0

. 6oin so ives us,

6x2+x12>0

'ext, we shall look into the correspondin equation,6x

2+x12=0

and factor it out to find its eroes as follows,(3x4 ) (2x+3 )=0

3x4=0 or 2x+3=0

x=4

3 orx=

32

'ow, these values divide the real number line into three intervals. +e shall plu in test values

within these intervals to see what interval satisfies the inequality. +e shall construct the sin

chart as follows,

23

20

4

32

3x4 +

2x+3 + +

(3x4 ) (2x+3) +

+

*ookin at the sins, clearly we see that the product is positive for the intervals ( ,32 )

and ( 43 , ) . 7eferrin to the inequality, the solution is iven by,

( ,32 )(4

3, )


6/26

4b5 In this problem, we shall use the properties of absolute value inequalities, specifically#

If |x|c , c 0 , then c x c .

Applyin this property and isolatin for x , we have as follows,

5 32x 5

82x 2

4 x 1

&herefore, the solution in interval notation is iven by,

[1,4 ]

4c5 "ince the expression is already factored, we can immediately solve for the critical values and

construct a sin chart as follows,

431 012

34

12 3

x + + +

x + + +

x+3 + + + +

4x3 + +

x2 +

x2 (x+3)( 4x3) (x2 )

+ + +


7/26

%y lookin at the sins, clearly we see that the product is neative for the intervals (,3 ]

and ( 34 , 2) . 7eferrin to the inequality, the solution is iven by,

(,3 ]( 34 ,2)

4d5 In this problem, we shall use the properties of absolute value inequalities, specifically#

If |x|c , c>0 , then xc .

Applyin this property, we et further two inequalities as follows,

2x+1x3

4 or

2x+1

x3 4

2x+1x3

+4 0 or

2x+1x3

4 0

6x11x3

0 or

2x+13x3

0

*ookin at their correspondin equations,

6x11x3 =0 or

2x+13x3 =0

and notin that they are already factored, we construct the sin chart for each as follows,

011

62 3 4 0 3 4

13

27


8/26

6x11 + + 2x+13 + +

x3 + x3 + +

6x11x3

+

+

2x+13x3

+

%ased from the sin chart, the first inequality is satisfied on the interval [116 ,3) , while the

second one is satisfied on the interval (3, 132] . &herefore, by combinin these results, we et,

[116 ,3)(3, 132]

. "ketch the reion iven by the system# {5x+3y


9/26

"imilarly, the reion defined by the second inequality is the open half-plane below the line y=2 .

2sin the information above, we raph the reion defined by the system of inequalities.

8raph#

i. /. &he reion defined by the inequality is the reion with darkest shadin.

9. 4a5 "ketch the reion iven by the system#

{

3xy+3 04x+5y34 0

x 0

y 2

4b5 ind the minimum value of z=4x+7y in the reion described in 4a5.

"olution#


10/26

4a5 +e rewrite the first two inequalities in slope-intercept form and use &heorem / to determine the

reion they define.

y 3x+3 and y 4

5x+

34

5

&hus, the reion defined by the first inequality is the half-plane below the line y=3

x+3

and

the reion defined by the second inequality is the half-plane below the liney=

45

x+34

5 .

"imilarly, the reion defined by the third inequality is the half-plane on the riht of the line

x=0 and the reion defined by the fourth inequality is the half-plane above the line y=2

.

2sin the information above, we raph the reion defined by the system of inequalities.

8raph#

i. . &he reion defined by the inequality is the reion with darkest shadin.

4b5 +e use &heorem to reduce the number of possible points that minimies z=4x+7y by

lookin at the vertices of the reion. +e shall test for the minimum amon the four vertices,

namely, (1,6) , (6,2) , (0,2) and (0, 3) as follows,


11/26

for (1,6 ) , z=4 (1 )+7 (6 )=46 ,

for (6,2 ) , z=4 (6 )+7 (2 )=38 ,

for (0,2 ) , z=4 (0 )+7 (2 )=14 , and

for (0,3 ) , z=4 (0 )+7 (3 )=21 .

&herefore the minimum value is at the point (0,2 ) with the minimum value,

z=14


12/26

:. 2se matrix inversion to solve the system# {4x2y=35x+y=2"olution#

rom the iven problem above, we set up the coefficient matrix A , the column vector of variables

x and the column vector of constantsb .

A=[4 25 1]x=[xy ]b=[ 32]+ith these, we can represent the system of linear equations as a matrix equation like so,

Ax=b

In the equation above, we are aimin to solve for the column vector of variables x . )ultiplyin the

equation with the inverse of matrix A , denoted by A1

, we et,

x=A1 b

In order to find x , we must first find A1

and multiply it with b . or a 2 by 2 matrix

A ,

A=[a bc d ]its inverse is iven by,

A1=

1

adbc[ d bc a]

Applyin the formula above for matrix A and multiplyin it with the column vector b , we have,

x= 1

(4 ) (1 ) (2 ) (5 )[ 1 25 4 ][ 32]


13/26

[xy]= 114 [123 ]=[1/1423 /14 ]&herefore,

x=114

, y=23

14

;. 2se elementary row operations to solve the system# {2x3y+5z=0

4x2z=62y+3z=3

"olution#

rom the iven problem above, we set up the aumented matrix like so,

[ 2 3 54 0 20 2 3| 0

6

3]rom then, we shall perform elementary row operations 48aussian elimination method5 until the matrix

is reduced to a trianular matrix.

irst, we replace row 2 with row 2 minus row 1 multiplied by 2 .

R2

r22 r

1[ 2 3 50 6 120 2 3| 0

6

3 ]'ext, we replace row 3 with row 2 plus row 3 multiplied by 3 .

R3

r2+3 r

3[ 2 3 50 6 120 0 3 | 0

6

3]


14/26

'otice that the coefficient matrix is now a trianular matrix. +e revert back to equations based from the

reduced aumented matrix above. &hen we solve the variables alebraically as follows,

2x3y+5z=0 4/5

6y12z=6 45

3z=3 495

s 7ule to solve the system# {2x+3y3z=1x+2y5z=24x+5y2z=4

"olution#

rom the iven problem, we set up the determinants D ,Dx ,

Dy , andDz as follows,

D=|2 3 31 2 54 5 2|Dx=|

1 3 32 2 5

4 5 2|D y=|2 1 31 2 54 4 2|Dz=|

2 3 1

1 2 24 5 4|

'ext, we compute the 9 by 9 determinants above usin the 7ule of "arrus.


15/26

D=(2 ) (2 ) (2 )+(3 ) (5 ) ( 4 )+(3 ) (1 ) (5 )(3 ) (2 ) (4 )(3 ) (1 ) (2 )(2 ) (5 ) (5 )=3

Dx= (1 ) (2 ) (2 )+(3) (5) ( 4 )+(3 ) (2 ) (5)(3 ) (2 ) (4 )(3 ) (2 ) (2 )(1) (5 ) (5 )=3

Dy=(2 ) (2) (2 )+ (1 ) (5 ) (4 )+ (3 ) (1 ) (4 )(3) (2 ) (4 )(1 ) (1 ) (2 )(2 ) (5 ) (4 )=6

Dz=(2) (2 ) (4 )+(3 ) (2 ) (4 )+(1 ) (1 ) (5) (1 ) (2) (4 )(3 ) (1 ) ( 4 )(2 ) (2) (5 )=3

&he values for x , y , and z are iven as follows,

x=Dx

D=

3

3=1

y=Dy

D=

63

=2

z=Dz

D=

33

=1

&o summarie,

x=1,y=2,z=1

Exam 2

/. ind all the eroes of the followin polynomials

4a5 P (x )=2x4+11x35x256x+48


16/26

4b5 P (x )=3x314x2+47x26

"olution#

4a5 %y applyin the 7ational ?eroes &heorem, the possible candidate roots for

P (x )is reduced to

a list as follows,

p= 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

q=1, 2

p

q= 1,

1

2, 2, 3,

3

2, 4, 6, 8, 12, 16, 24, 48

%y applyin the actor &heorem, we shall evaluate P (x ) with these values and look for values

that satisfies P (x )=0 . %y trial and error, one will find these values,

P (1)=0 , so x=1 is a ero,

P (4 )=0 , so x=4 is a ero and

P(32 )=

0, so x=

32 is also a ero.

"ince P (x ) is a polynomial of deree 4 and no more candidate roots satisfies the actor

&heorem, then the remainin ero of P (x ) could be a repeated, irrational or imainary root.

6ividin P (x ) by the factors we found so far usin synthetic division successively,

1 2 11 5 56 48

2 13 8 48

4 2 13 8 48 0

8 20 48


17/26

3

2

2 5 12 0

3 12

2 8 0

which means that P (x )= (x+4 )2 (2x3 ) (x1) , showin that it has a repeated root.

"o, the eroes of P (x ) are,

4,1,3

2

4b5 %y applyin the 7ational ?eroes &heorem, the possible candidate roots for P (x ) is reduced to

a list as follows,

p= 1, 2, 13, 26

q=1, 3

p

q= 1,

1

3, 2,

2

3, 13,

13

3, 26,

26

3

%y applyin the actor &heorem, we shall evaluate P (x ) with these values and look for values

that satisfies P (x )=0 . %y trial and error, one will find,

P( 23 )=0 , so x=2

3 is a ero.

"ince

P (x )is a polynomial of deree

3

and no more candidate roots satisfies the actor

&heorem, then P (x ) must have at least two more repeated, irrational or imainary roots.

6ividin P (x ) with (x23 ) usin synthetic division, we et,


18/26

2

3

3 14 47 26

2 8 26

3 12 39 0

which means that P (x )= (3x2 )(x24x+13) .

%y examinin the quadratic factor (x24x+13 ) , its discriminant, b

24 ac=36 is clearly

less than 0 , so its roots are imainary. Applyin the quadratic formula, we have,

x=(4 ) (4 )

24 (1 ) (13)

2(1 ) =2 3 i

"o P (x )= (3x2 ) (x+2+3 i) (x+23 i ) and its eroes are,

2

3, 2+3i , 23 i

. 6etermine whether x=1 is a ero of f(x )=2x1433x

84+6x

17+10 .

"olution#

%y actor &heorem, (xr ) is a factor of P (x ) , if and only if P (r )=0 .


19/26

"ince f(1 )0 , x=1 is not a ero of f(x ) .

9. ind the first term of an arithmetic proression whose second term is 1 and whose eihth

term is 8 .

"olution#

2sin the formula for the n th term of an arithmetic proression,

an=a1+(n1 )d

and lettin

a2=1

and

a8=8

, we have a system of two equations in two unknowns like so,

{a2=a1+d=1a8=a

1+7 d=8

)ultiplyin the first equation with 7 and addin to it the second equation, we et,

6 a1=15

&herefore, the first term is iven by,

a1=5

2


20/26

:. Insert 4 arithmetic means between 20 and 7 .

"olution#

+e are lookin for numbersc

1 throuhc

4 which satisfies the arithmetic proression

a , c1

, c2

, c3

, c4

, b, @, where a=20 and b=7 . %y settin

a1=a

,a

2=c

1 , @,

a5=c

4 , anda

6=b

, our oal is to find out the termsa

2 throuha5 . 2sin the formula for the

n th term of an arithmetic proression,

an=a1+(n1 )d

we shall look first for the common difference,d

. "ubstitutin in the appropriate values, we have,

7=20+5 dd=13

5

Aain, usin the same formula, we shall look for the termsa

2 throuha

5 like so,

a2=a1+d=20+

13

5=87

5

a3=a

1+2 d=20+2( 135)=

74

5

a4=a

1+3 d=20+3( 135)=

61

5

a5=a

1+4 d=20+4 ( 135)=485

'ow, sincea

2=c

1 , @,a

5=c

4 , the four arithmetic means between 20 and 7 are,

875

,74

5,

61

5,

48

5


21/26

;. ind the second term of the eometric proression whose fifth term is27

8 if the common ratio

is32 .

"olution#

2sin the formula for the n th term of a eometric proression,

an=a1 rn1

the second term, a2 is iven by,

a2=a

1r

4/5

"imilarly, the fifth term is iven by,

a5=a1 r4

"olvin fora

1 and substitutin it into equation 4/5, we have as follows,

a1=

a5

r4


22/26

a2=a

1r=( a5r4 )r=

a5

r3

"ubstitutin in the values, we et,

a2=

27

8

(32 )3=( 278)(

8

27 )

a2=1

=. ind the sum of the infinite eometric series whose second term is8

15 and whose third term

is16

75 .

"olution#

irst, we need to find out the first term

a1

and the common ratio

r

of the eometric proression.

2sin the recursive formula for the n th term of a eometric proression,

an=an1 r


23/26

the third terma

3 is iven by,

a3=a

2r

"olvin for r and substitutin in the appropriate values,

r=a

3

a2

=

1675

8

15

=2

5

Aain, by the recursive formula for the n th term of a eometric proression,a

1 is iven by,

a1=

a2

r=

815

2

5

=4

3

'otice that |r|


24/26

4a5 In how many ways could they choose a president, a secretary and a treasurer!4b5 In how many ways could they choose a president, a secretary and a treasurer if the secretary

is a boy and the treasurer is a irl!

"olution#

4a5 &he order at which we choose a president, a secretary and a treasurer is important. All in all,

there are 12+9=21 candidates to place for the said positions. "ince, no sinle candidate

is allowed to take up two different positions and the order of which candidate we place on the

said positions matters, then by undamental Principle of 1ountin, the number of ways they

could choose three officers from twenty-one candidates is iven by,

21 2019=7980 ways.

P " &

4b5 "ince the position for the secretary or the treasurer is ender specific, we should choose a

secretary or a treasurer first before a president. Aain, the order at which we choose matters,

so upon fillin the secretary position first, which requires a boy, allows us to choose 12

candidates. "imilarly, choosin for a treasurer, which requires a irl, allows us to choose

9 candidates. *astly, for the president, since no sinle candidate is allowed to take up two

different positions and we have chosen two candidates already, there are only 212=19

candidates left to choose. %y the undamental Principle of 1ountin, the number of waysthey could choose three officers takin into consideration the constraints is iven by,

12 919=2052 ways.

" & P

B. (ow many permutations of the letters of the word "


25/26

3. A shelf contains 4 different math books and 5 different enlish books.

4a5 In how many ways could the books be arraned on the shelf if the math books are kepttoether and the enlish books are kept toether!

4b5 In how many ways could the books be arraned on the shelf if the math books are kept

toether!

"olution#

4a5 +e must keep the enlish and math books toether, so we must take the cases separately

whenever the enlish or math books were placed toether first. &akin the case when the

enlish books were placed toether first, then by the undamental Principle of 1ountin, the

number of ways we could arrane the books is iven by,

5 4 3 2 1

E

4 3 2 1

M

=2880ways.

Dn the other hand, if the math books were placed toether first, then by the undamental

Principle of 1ountin, the number of ways we could arrane the books is iven by,

4 3 21

M

5 4 3 2 1

E

=2880ways.

%y the rule of sum, the total number of ways the books can be arraned takin the constraints

into consideration is iven by

5 !4 !+4 !5 !=2880+2880=5760 ways.

4b5 +e shall treat the math books toether as a sinle obect. &he number of ways the five

enlish books and the math books treated as a sinle obect can be arraned is iven by,

6 5 4 3 2 1=6 !=720 ways. 4/5

or every arranement above, the math books can be arraned in,

4 32 1=4 !=24 ways. 45

%y the undamental Principle of 1ountin, the total number of ways the books can be

arraned considerin the constraints is simply the product of 4/5 and 45, namely,


26/26

6 ! 4 !=6 5 4 3 21 4 3 21=17280 ways.

Documents

Problems in Advanced Algebra