Chapter 6 5 Glencoe Precalculus
6-1 Study Guide and Intervention Multivariable Linear Systems and Row Operations
Gaussian Elimination You can solve a system of linear equations using matrices. Solving a system by transforming it
into an equivalent system is called Gaussian elimination. First, create the augmented matrix. Then use elementary row
operations to transform the matrix so that it is in row-echelon form. Then write the corresponding system of equations
and use substitution to solve the system..
Example: Solve the system of equations using Gaussian elimination with matrices.
x β 2y + z = β1
2x + y β 3z = β7
3x β y + 2z = 0
Step 1 Write the augmented matrix
[1 β2 12 1 β33 β1 2
|β1β7
0]
Step 2 Apply elementary row operations to obtain a rowβechelon form of the matrix.
a. b. c.
π 2 β 2π 1 β [1 β2 10 5 β53 β1 2
|β1β5
0]
π 3 β 3π 1 β[1 β2 10 5 β50 5 β1
|β1β5
3]
1
5π 2 β [
1 β2 10 1 β10 5 β1
|β1β1
3]
d. e.
π 3 β 5π 2 β[1 β2 10 1 β10 0 4
|β1β1
8]
1
4π 3 β
[1 β2 10 1 β10 0 1
|β1β1
2]
Step 3 Write the corresponding system of equations and use substitution to solve the system.
x β 2y + z = β1
y β z = β1
z = 2
The solution of the system is x = β1, y = 1, and z = 2 or (β1, 1, 2).
Exercises Solve each system of equations using Gaussian elimination with matrices.
1. β2x β y + z = 0 2. 5x β y = β13 3. β4x β y β z = β11
x + 2y β z = β3 β3x + 2y β z = 8 x β z = 2
3x + y β 2z = β1 x β 4y + z = β10 2y + 4z = 0
(2, β1, 3) (β2, 3, 4) (3, β2, 1)
Chapter 6 6 Glencoe Precalculus
6-1 Study Guide and Intervention (continued)
Multivariable Linear Systems and Row Operations
Gauss-Jordan Elimination If you continue to apply elementary row operations to the row-
echelon form of any augmented matrix, you can obtain a matrix in which every column has one
element equal to 1 and the remaining elements equal to 0. This is called the reduced row-
echelon form of the matrix. Solving a system by transforming an augmented matrix so that it is
in reduced rowβechelon form is called GaussβJordan elimination..
[1 0 00 1 00 0 1
| πππ
]
Example : Solve the system of equations.
x β 2y + z = 15
β2x β y + 2z = β1
βx + y = β9
Write the augmented matrix. Apply elementary row operations to obtain a rowβechelon form. Then apply elementary row operations to obtain zeros above the leading 1s in each row.
Augmented Matrix
[1 β2 1
β2 β1 2
β1 1 0
|15
β1β9
] 2π 1 + π 2 β [1 β2 1
0 β5 4
β1 1 0
|15
29
β9]
π 1 + π 3 β[1 β2 1
0 β5 4
0 β1 1
|15
29
6
]
Rowβechelon form
π 2 β 6π 3 β [1 β2 1
0 1 β20 β1 1
|15
β76
]
π 2 + π 3 β[1 β2 1
0 1 β20 0 β1
|15
β7β1
]
β π 3 β[1 β2 1
0 1 β20 0 1
|15
β71
]
Reduced row-echelon form
π 2 + 2π 3 β [1 β2 1
0 1 0
0 0 1
|15
β51
]
π 1 + 2π 2 β[1 0 1
0 1 0
0 0 1
|5
β51
]
π 1 β π 3 β[1 0 0
0 1 0
0 0 1
|4
β51
]
The solution of the system is x = 4, y = β5, and z = 1 or (4, β5, 1).
Exercises
Solve each system of equations using Gaussian or GaussβJordan elimination.
1. 3x β 2y + z = β22 2. x β 4z = 6 3. β2x β y β z = 1
β4x + z = 31 β2y + 3z = β2 βx + 3y β 2z = 24
2x β 5y = β24 2x β 5y = 6 4x + 2y + z = 2
(β7, 2, 3) (β2, β2, β2) (β1, 5, β4)
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 7 Glencoe Precalculus
6-1 Practice Multivariable Linear Systems and Row Operations
Write each system of equations in triangular form using Gaussian elimination. Then solve the system.
1. [1 β1
β3 2|β1232
] 2. [2 3 1
β3 β1 4β1 5 β1
|β23β5
β19] 3. [
β5 β3 00 β2 64 0 β7
|β2242]
Write the augmented matrix for each system of linear equations.
4. 5x β 2y = 14 5. 3x + 4y + 7z = β8 6. β4x β 2y β z = 5
β3x + y = β7 β2x β 3y + z = 6 2x β z = 8
5x β 2y + z = 4 y β 2z = β4
Solve each system of equations using GaussβJordan elimination.
7. β4x β 2y = β6 8. β2x β 5y + z = 6 9. 8x β y + 3z = β38
x + 3y = β11 3x + 2y β 4z = β1 β2x + 5y β 4z = 32
5x β y + 2z = β6 x β y + z = β9
10. FRUIT Three customers bought fruit at Michaelβs Groceries. The table shows the amount of fruit bought by each
person. Write and solve a system of equations to determine the price of each type of fruit.
Name Apples Oranges Pears Total Cost ($)
Rosario 5 4 3 13.50
Lindsay 7 2 4 14.20
Edwin 3 8 2 15.30
(β8, 4) (β2, β5, β4) (4, β6, 2)
[π βπ
βπ π|ππβπ
] [π π π
βπ βπ ππ βπ π
|βπππ] [
βπ βπ βππ π βππ π βπ
|ππ
βπ
]
(4, β5) (β1, β1, β1) (β3, 2, β4)
[π π ππ π ππ π π
|ππ. ππππ. ππππ. ππ
]; (1.1, 1.25, 1); apples: $1.10, oranges: $1.25, pears: $1
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 11 Glencoe Precalculus
6-2 Study Guide and Intervention Matrix Multiplication, Inverses, and Determinants
Multiply Matrices To multiply matrix A by matrix B, the number of columns in A must be equal to the number of rows
in B. If A has dimensions m Γ r and B has dimensions r Γ n, their product, AB, is an m Γ n matrix. If the number of columns in A does not equal the number of rows in B, the matrices cannot be multiplied.
[π ππ π
]β [π ππ β
] =[ππ + ππ ππ + πβππ + ππ ππ + πβ
]
Example: Use matrices A = [ π βπβπ π
] and B =[βπ π πβπ π βπ
]to find AB, if possible.
AB = [4 β2
β1 3]β [
β1 2 3β2 4 β1
]
A is a 2 Γ 2 matrix and B is a 2 Γ 3 matrix. Because the number of columns for A is equal to the number of rows for B,
the product AB exists.
To find the first entry in AB, write the sum of the products of the entries in row 1 of A and in column 1 of B.
[4 β2
β1 3]β [
β1 2 3β2 4 β1
]= [π(β π) + (β π)(β π)]
Follow this same procedure to find the entry for row 1, column 2 of AB.
[4 β2
β1 3]β [
β1 2 3β2 4 β1
]= [π(β π) + (β π)(β π) π(π) + (β π)(π)]
Continue multiplying each row by each column to find the sum for each entry.
[4 β2
β1 3]β [
β1 2 3β2 4 β1
]= [4(β1) + (β2)(β2) 4(2) + (β2)(4) 4(3) + (β2)(β1)
(β1)(β1) + 3(β2) (β1)(2) + 3(4) (β1)(3) + 3(β1)]
Then simplify each sum.
[4 β2
β1 3]β [
β1 2 3β2 4 β1
]= [0 0 14
β5 10 β6]
Exercises
Find AB and BA, if possible.
1. A = [2 4
β3 β1], B = [
β1 50 β2
] 2. A = [β1 3β3 2
], B = [β2 4 0β3 β1 2
]
AB =[βπ π
π βππ]; BA = [
βππ βππ π
] AB =[βπ βπ π
π βππ π]; BA is undefined.
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 12 Glencoe Precalculus
6-2 Study Guide and Intervention(continued)
Matrix Multiplication, Inverses, and Determinants
Inverses and Determinants The identity matrix is an n Γ n matrix consisting of all 1s on its main diagonal, from
upper left to lower right, and 0s for all other elements. LetπΌπbe the identity matrix of order n and let A be an n Γ n matrix.
If there exists a matrix B such that AB = BA = πΌπ , then B is called the inverse of A and is written as π΄β1. If a matrixhas an
inverse, it is invertible. The determinant of a 2 Γ 2 matrix can be used to determine whether or not a matrix is invertible.
If A =[π ππ π
], det(A) = ad βcb. If ad βcb β 0, then π΄β1=1
ππ βππ[
π βπβπ π
].
Determine whether A = [π βπ
βπ π] and B = [
π ππ π
]are inverse matrices.
If A and B are inverse matrices, then AB = BA = I.
AB = [7 β4
β5 3]β [
3 45 7
] = [7(3) + (β4)(5) 7(4) + (β4)(7)
β5(3) + 3(5) β5(4) + 3(7)] or [
1 00 1
]
BA = [3 45 7
]β [7 β4
β5 3] = [
3(7) + 4(β5) 3(β4) + 4(3)
5(7) + 7(β5) 5(β4) + 7(3)] or [
1 00 1
]
Because AB = BA = I, B = π΄β1and A = π΅β1.
Find the determinant of A = [π βππ βπ
].Then find π¨βπ, if it exists.
det(A) = |2 β23 β6
| Aβ1 = β1
6[β6 2β3 2
]
= 2(β6) β3(β2) or β6 = [1 β
1
31
2β
1
3
]
Since det(A) β 0, A is invertible.
Exercises
Determine whether A and B are inverse matrices. Explain your reasoning.
1. A=[11 5
2 1], B = [
1 β5
β2 11] 2.A = [
3 2
4 1], B = [
1 5
4 3]
3. Find the determinant of A = [5 β1
β10 2]. Then find π΄β1, if it exists.
4. Find the determinant of A = [3 2
1 β1]. Then find π΄β1, if it exists.
yes; AB = BA = I2 no; AB β BA β I2
det(A) = 0; Aβ1 does not exist.
det(A) = β5, Aβ1 =[π. π π. ππ. π βπ. π
]
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 13 Glencoe Precalculus
6-2 Practice Matrix Multiplication, Inverses, and Determinants
Find AB and BA, if possible.
1. A = [β1 6 0
3 β2 1] , B = [
2 β4β1 3
] 2. A = [3 0
β1 2], B = [
3 5β2 0
]
3. GOLF The number of golf clubs manufactured daily by two different companies is shown, as well as the selling price of each type of club. Use this information to determine which companyβs daily production has the highest retail value.
How much greater is the value?
Company Club Type and Quantity
1-Wood 3-Wood 5-Wood Putter
A 600 520 310 300
B 210 400 450 400
Club Club Value ($)
1-Wood 210
3-Wood 170
5-Wood 150
Putter 120
Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the
augmented matrix to solve for X.
4. π₯1 β 2π₯2 + 3π₯3 = 4 5. 2π₯1 + π₯2 + 2π₯3 = 11
5π₯1 + 3π₯2 β π₯3 = 13 β5π₯1 β π₯2 + 4π₯3 = 1
4π₯1 β π₯2 + 4π₯3 = 11 3π₯1 β 2π₯2 + 8π₯3 = 28
Determine whether A and B are inverse matrices.
6. A = [1 21 3
], B = [3 β2
β1 1] 7. A = [
5 β24 β3
], B = [β1 0
2 β8]
Find the determinant of each matrix. Then find its inverse, if it exists.
8. [6 52 2
] 9. [β2 4
3 β6]
Evaluate.
A = [β1 5
3 0] B = [
β4 2 β10 β5 3
] C = [β1 0 β4
3 β2 1]
10. AB + C 11. A(B β C)
AB is undefined; BA = [βππ ππ βπ
ππ βππ π] AB = [
π ππβπ βπ
]; BA = [π ππ
βπ π]
Company A; $69,300
[π βπ ππ π βππ βπ π
] β [
ππ
ππ
ππ
] = [π
ππππ
] (4, β3, β2) [π π π
βπ βπ ππ βπ π
] β [
ππ
ππ
ππ
] = [πππ
ππ] (2, 1, 3)
yes no
2; [π βπ. π
βπ π] 0; singular
[π βππ ππ
βπ π βπ] [
βππ βππ πβπ π π
]
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 16 Glencoe Precalculus
6-3 Study Guide and Intervention Solving Linear Systems Using Inverses and Cramerβs Rule
Use Inverse Matrices A square system has the same number of equations as variables. If a square matrix has an
inverse, the system has one unique solution.
Example : Use an inverse matrix to solve each system of equations, if possible.
a. 3x β 7y = β16
βx + 2y = 8
Write the system in matrix form.
A β X = B
[3 β7
β1 2] β [
π₯π¦] = [
β168
]
Use the formula for an inverse of a 2 Γ 2 matrix to find
the inverse π΄β1.
π΄β1 = 1
ππ β ππ [
π βπβπ π
]
= 1
(2)(3) β (β7)(β1) [
2 71 3
]
Multiply π΄β1 by B to solve the system.
X = π΄β1 β B
= [β2 β7β1 β3
] β [β16
8] = [
β24β8
]
So, the solution of the system is (β24, β8).
b. β2x + y + z = 0
x + 2z = 9
x β 2y β 9z = β31
Write the system in matrix form.
A β X = B
[β2 1 1
1 0 21 β2 β9
] β [π₯π¦π§
] = [09
β31]
Use a graphing calculator to find π΄β1.
Multiply π΄β1 by B to solve the system.
X = π΄β1 β B
= [4 7 2
11 17 5β2 β3 β1
] β [09
β3] = [
1β2
4]
So, the solution of the system is (1, β2, 4).
Exercises
Use an inverse matrix to solve each system of equations, if possible.
1. β2x + 5y = 24 2. x β y + 2z = 5
3x β y = β10 x β z = β4
3x + 2y + z = 0
3. 3x + y = 7 4. x + y β z = β5
β2x β 5y = 43 2x β 3y + 2z = 20
y + 4z = 18
(β2, 4) (β1, 0, 3) (6, β11) (2, β2, 5)
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 17 Glencoe Precalculus
6-3 Study Guide and Intervention (continued)
Solving Linear Systems Using Inverses and Cramerβs Rule
Use Cramerβs Rule Another method, known as Cramerβs Rule, can be used to solve a square system of equations.
Let A be the coefficient matrix of a system of n linear equations in n variables given by AX = B. If det(A) β 0, then the
unique solution of the system is given by
π₯1 = |π΄1|
|π΄|, π₯2 =
|π΄2|
|π΄|, π₯3 =
|π΄3|
|π΄|β¦, π₯π =
|π΄π|
|π΄|,
where Ai is the matrix obtained by replacing the ith column of A with the column of constants B. If det(A) = 0, then AX = B has either no solution or infinitely many solutions.
Example: Use Cramerβs Rule to find the solution of the system of linear equations, if a unique solution exists.
β2ππ + ππ = β7
5ππ β 2ππ = 17
The coefficient matrix is A = [β2 1
5 β2]. Calculate the determinant of A.
|π΄| = |β2 1
5 β2| = (β2) (β2) β 5(1) or β1
Because the determinant of A does not equal zero, you can apply Cramerβs Rule.
π₯1 = |π΄1|
|π΄| =
|β7 117 β2
|
|β1| =
β7(β2) β 17(1)
β1 =
β3
β1 or 3
π₯2 = |π΄2|
|π΄| =
|β2 β7
5 17|
β1 =
β2(17) β 5(β7)
β1 =
1
β1 or β1
Therefore, the solution is π₯1 = 3 and π₯2 = β1 or (3, β1).
Exercises
Use Cramerβs Rule to find the solution of each system of linear equations, if a unique solution exists.
1. x β 2y = β5 2. 3x β 3y = β18
β2x β 5y = β8 βx + 4y = 9
3. 3x + y = 21 4. β2x β 4y = 2
βx + 2y = 14 x + 3y = β3
(β1, 2) (β5, 1) (4, 9) (3, β2)
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 18 Glencoe Precalculus
6-3 Practice Solving Linear Systems Using Inverses and Cramerβs Rule
Use an inverse matrix to solve each system of equations, if possible.
1. 4x β 7y = 30 2. β2x β 8y = β36
β6x + 2y = β11 4x + 3y = 7
3. x β 2y + 7z = β33 4. x + y β 2z = 5
β4x + 5y β z = 18 x + 2y + z = 8
5x β 3y = β11 2x + 3y β z = 1
5. TELEVISION During the summer, Manuel watches television M hours per day, Monday through Friday. Harry
watches television H hours per day, Friday and Saturday. Ellen watches television E hours per day, Friday through
Sunday. Altogether, they watch television 37 hours each week. On Fridays, they watch a total of 11 hours of television. If the number of hours Ellen spends watching television on any given day is twice the number of hours that Manuel
spends watching television on any given day, how many hours of television does each of them watch each day?
Use Cramerβs Rule to find the solution of each system of linear equations, if a unique solution exists.
6. β4x β 5y = 1 7. x + y + z = 8
β2x β 3y = β1 3x β z = β22
y + 2z = 20
8. PAPER ROUTE Payton, Santiago, and Queisha each have a paper route. Payton delivers 5 times as many papers as
Santiago. Santiago delivers twice as many papers as Queisha. If 20 papers were added to Paytonβs route, he would then deliver four times the total number of papers that Santiago and Queisha deliver. How many papers does each person
deliver?
(π
π, βπ) (β2, 5)
(β1, 2, β4) no solution
Manuel: 3 hours, Harry: 2 hours, Ellen: 6 hours
(β 4, 3)
(β5, 6, 7) Payton delivers 100; Santiago delivers 20; Queisha delivers 10
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 22 Glencoe Precalculus
6-4 Study Guide and Intervention Partial Fractions
Linear Factors The function g(x) shown below can be written as the sum of two fractions with denominators that are
linear factors of the original denominator.
g(x) = 3π₯ β 1
2π₯2 β 3π₯ +1 =
2
π₯ β 1 +
β1
2π₯ β 1
Each fraction in the sum is a partial fraction. The sum of these partial fractions make up the partial fraction
decomposition of the original rational function. If the denominator of a rational expression contains a repeated linear
factor, the partial fraction decomposition must include a partial fraction with its own constant numerator for each power of this factor.
Example : Find the partial fraction decomposition of π + ππ
ππ β ππ β π.
Rewrite the equation as partial fractions with constant numerators, A and B, and denominators that are the linear factors of
the original denominator.
π₯ +11
π₯2β 3π₯β4 =
π΄
π₯ β 4 +
π΅
π₯ +1 Form a partial fraction decomposition.
x + 11 = A(x + 1) + B(x β 4) Multiply each side by the LCD, π₯2 β 3x β 4.
x + 11 = Ax + A + Bx β 4B Distributive Property
1x + 11 = (A + B)x + (A + (β4B)) Group like terms.
Equate the coefficients on the left and right side of the equation to obtain a system of two equations. To solve the system,
write it in matrix form CX = D and solve for X.
A + B = 1
A + (β4B) = 11 βΆ
C β X = D
[1 11 β4
] β [π΄π΅
] = [1
11]
Solving for X yields A = 3 and B = β2. Therefore π₯ +11
π₯2β 3π₯β4 =
3
x β 4 +
β2
x +1.
Exercises
Find the partial fraction decomposition of each rational expression.
1. 5π₯ β 34
π₯2β π₯ β12 2.
β7π₯ + 13
π₯2 β 5π₯ β14
3. π₯2 + 1
2π₯(π₯ β 1)2 4. 5π₯2 β π₯ β 1
π₯2(π₯ β 1)
π
π + π +
βπ
π β π
βπ
π β π +
βπ
π + π
π
ππ +
π
(πβπ)π π
π +
π
ππ + π
πβπ
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 23 Glencoe Precalculus
6-4 Study Guide and Intervention (continued)
Partial Fractions
Irreducible Quadratic Factors Not all rational expressions can be written as the sum of partial fractions using only
linear factors in the denominator. If the denominator of a rational expression contains an irreducible quadratic factor, the partial fraction decomposition must include a partial fraction with a linear numerator of the form Bx + C
for each power of this factor.
Example : Find the partial fraction decomposition of πππ β πππ β ππππ + ππ + π
π(ππ β π)π
This expression is proper. The denominator has one linear factor and one irreducible factor of multiplicity 2.
πππ β πππ β ππππ + ππ + π
π(ππβ π)π = π¨
π +
π©π + πͺ
ππ β π +
π«π + π¬
(ππ β π)π
4π₯4 β 2π₯3 β13π₯2 + 7x + 9 = A(π₯2 β 3)2 + (Bx + C)x(π₯2 β 3) + (Dx + E)x
4π₯4 β 2π₯3 β13π₯2 + 7x + 9 = Aπ₯4 + Bπ₯4 + Cπ₯3β 6Aπ₯2 β 3Bπ₯2 + Dπ₯2 β 3Cx + Ex + 9A
4π₯4 β 2π₯3 β13π₯2 + 7x + 9 = (A + B) π₯4 + Cπ₯3 + (β6A β 3B + D) π₯2 + (β3C + E)x + 9A
Write and solve the system of equations obtained by equating coefficients.
A + B = 4 A = 1
C = β2 B = 3
β6A β 3B + D = β13 β C = β2
β3C + E = 7 D = 2
9A = 9 E = 1
Therefore, 4π₯4 β 2π₯3 β 13π₯2 + 7π₯ + 9
π₯(π₯2 β 3)2 = 1
π₯ +
3π₯ + 2
π₯2 β 3 +
2π₯ + 1
(π₯2 β 3)2
Exercises
Find the partial fraction decomposition of each rational expression.
1. 5
π₯3+ 5π₯ 2.
3π₯3 β 2π₯2β 8π₯ + 5
(π₯2 β 3)2
3. 2π₯3 + π₯ + 3
(π₯2 + 1)2 4. π₯3 + 2π₯2 + 2
(π₯2 + 1)2
π
π β
π
ππ+π
ππ β π
ππβπ +
π β π
(ππβπ)π
ππ
ππ+π +
πβπ
(ππ+π)π
π + π
ππ+ π +
βπ
(ππ+π)π
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 24 Glencoe Precalculus
6-4 Practice Partial Fractions
Find the partial fraction decomposition of each rational expression.
1. 3π₯ β 7
π₯2 β 7π₯ + 12 2.
6π₯2 β 10π₯ β 2
π₯3 + π₯2 β 2π₯
3. 9π₯ + 15
π₯2 + 3π₯ + 2 4.
π₯
2π₯2 β 9π₯ + 9
Find the partial fraction decomposition of each improper rational expression.
5. 3π₯2 + 5π₯ + 2
π₯2 + 2π₯ 6.
β5π₯2 β 11π₯ + 54
π₯2 + 2π₯ β 8
7. 6π₯2 + 17π₯ + 2
π₯2 + π₯ 8.
β8π₯2 + 22π₯ β 10
(2π₯ β 3)π
Find the partial fraction decomposition of each rational expression with repeated factors.
9. β2π₯2 + 29π₯ β 100
π₯3 β 10π₯2 + 25π₯ 10.
5π₯4 β 7π₯3 β 12π₯2 + 6π₯ + 21
(π₯ β 3)(π₯2 β 2)2
11. 2π₯2 + 5
π₯3 + 6π₯2 + 9π₯ 12.
4π₯4 + 8π₯3 + 6π₯2 + 6π₯ + 5
(3π₯ + 2)(π₯2 + 1)2
13. GROWTH When working with exponential growth in calculus, it is often necessary to work with functions of the
form f(x) = 1
π₯(50 β π₯) and to decompose these functions into the sum of its partial fractions. Find the partial
decomposition of f(x).
π
π β π +
βπ
π β π
π
π +
βπ
π β π +
π
π + π
π
π + π +
π
π + π
π
π β π +
βπ
ππ β π
3 + π
π +
βπ
π + π β5 +
π
π β π +
βπ
π + π
6 + π
π +
π
π + π β2 β
π
ππ β π +
π
(ππ β π)π
βπ
π +
π
π β π +
βπ
(π β π)π π
π β π +
ππ β π
ππβπ +
π β π
(ππβπ)π
π
ππ +
ππ
π(π + π) +
βππ
π(π + π)π π
ππ + π +
π + π
ππ + π +
βπ
(ππ+π)π
π
ππ
π +
π
ππ
ππβπ or
π
πππ +
π
ππ(ππβπ)
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 27 Glencoe Precalculus
6-5 Study Guide and Intervention Linear Optimization
Linear Programming Linear programming is a process for finding a minimum or maximum value for a specific
quantity. The following steps can be used to solve a linear programming problem.
Step 1 Write an objective function and a list of constraints to model the situation.
Step 2 Graph the region corresponding to the solution of the system of constraints.
Step 3 Find the coordinates of the vertices of the region formed.
Step 4 Evaluate the objective function at each vertex to find the minimum or maximum.
Example: A leather company wants to add belts and wallets to its product line. Belts require 2 hours of cutting
time and 6 hours of sewing time. Wallets require 3 hours of cutting time and 3 hours of sewing time. The cutting
machine is available 12 hours a week and the sewing machine is available 18 hours per week. Belts will net $18 in
profit and wallets will net $12. How much of each product should be produced to achieve maximum profit?
Let x represent the number of belts and y represent the number of wallets.
The objective function is then given by f(x, y) = 18x + 12y.
Write the constraints.
x β₯ 0; y β₯ 0 Numbers of items cannot be negative.
2x + 3y β€ 12 Cutting time
6x + 3y β€ 18 Sewing time
Graph the system. The solution is the shaded region, including its boundary segments.
Find the coordinates of the four vertices by solving the system of boundary equations
for each point of intersection. The coordinates are (0, 0), (0, 4), (1.5, 3), and (3, 0).
Evaluate the objective function for each ordered pair.
Point f(x, y) = 18x + 12y Result
(0, 0) f(0, 0) = 18(0) + 12(0) 0
(0, 4) f(0, 4) = 18(0) + 12(4) 48
(1.5, 3) f(1.5, 3) = 18(1.5) + 12(3) 63 β Maximum
(3, 0) f(3, 0) = 18(3) + 12(0) 54
Since f is greatest at (1.5, 3), the company will maximize profit if it makes and sells 1.5 belts for every 3 wallets.
Exercises
Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur,
subject to the given constraints.
1. f(x, y) = 3x β 2y 2. f(x, y) = x + 2y
2x + y β€ 10 x + y β€ 4
x + 2y β€ 8 x + 3y β€ 6
x β₯ 0 x β₯ 0
y β₯ 0 y β₯ 0
max at (5, 0) = 15, max at (3, 1) = 5, min at (0, 4) = β8 min at (0, 0) = 0
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 28 Glencoe Precalculus
6-5 Study Guide and Intervention (continued)
Linear Optimization
No or Multiple Optimal Solutions Linear programming models can have one, multiple, or no optimal solutions.
If the graph of the objective function f to be optimized is coincident with one side of the region of feasible solutions, f has multiple optimal solutions. If the region does not form a polygon, but instead is unbounded, f may have no
minimum value or maximum value.
Example: Find the maximum value of the objective function f(x, y) = 6x + 3y and for what values of x and y it
occurs, subject to the following constraints.
2x + y β€ 8 y β€ 4 x β€ 3
x β₯ 0 y β₯ 0
Graph the region bounded by the given constraints. Find the value of the objective function f(x, y) = 6x + 3y at each vertex.
f(0, 0) = 6(0) + 3(0) or 0
f(0, 4) = 6(0) + 3(4) or 12
f(2, 4) = 6(2) + 3(4) or 24
f(3, 2) = 6(3) + 3(2) or 24
f(3, 0) = 6(3) + 3(0) or 18
Because f(x, y) = 24 at (2, 4) and (3, 2), the problem has multiple optimal solutions. An equation of the line through these
two vertices is y = β2x + 8. Therefore, f has a maximum value of 24 at every point on y = β2x + 8 for 2 β€ x β€ 3.
Exercises
Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur,
subject to the given constraints.
1. f(x, y) = 2x + y 2. f(x, y) = 3x + 6y
2x + y β€ 11 x - y β₯ β4
y β€ 5 x + 2y β€ 20
y β₯ 0 x β₯ 0
x β€ 5 x β€ 8
x β₯ 0 y β₯ 0
max. of 11 at every max. of 60 at every
point on y = β2x + 11 point on y = β π
πx + 10
for 3 β€ x β€ 5; for 4 β€ x β€ 8; min. at (0, 0) = 0 min. at (0, 0) = 0
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 6 29 Glencoe Precalculus
6-5 Practice Linear Optimization
Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they
occur, subject to the given constraints.
1. f(x, y) = 2x + 5y 2. f(x, y) = 4x + 3y
x β₯ 0 x β₯ 0
y β₯ 0 y β₯ 0
x + y β€ 7 2x + 3y β₯ 6
2x + 3y β€ 18 x + y β€ 8
3. f(x, y) = 2x β 3y 4. f(x, y) = 3x + 3y
x β₯ 0 x β₯ 0
x β€ 7 y β₯ 0
y β₯ 0 y β€ 8
y β€ 5 x + y β€ 10
x + 2y β₯ 14 3x + 2y β€ 24
5. SKATES A manufacturer produces roller skates and ice skates.
Manufacturer Information
Roller Skates Ice Skates Maximum Time Available
Assembling 5 minutes 4 minutes 200 minutes
Checking and Packaging 1 minute 4 minutes 120 minutes
Profit per Skate $40 $30
a. Write an objective function and list the constraints that model the given situation.
b. Sketch a graph of the region determined by the constraints from part a to find the
set of feasible solutions for the objective function.
c. How many roller skates and ice skates should be manufactured to maximize profit?
What is the maximum profit?
d. Describe why the company would choose a number of roller skates and ice skates
different from the answer in part c.
max. at (0, 6) = 30, max. at (8, 0) = 32, min. at (0, 0) = 0 min. at (0, 2) = 6
max. at (7, 3.5) = 3.5, max. of 30 at every point on min. at (4, 5) = β7 y = βx + 10 for 2 β€ x β€4,
min. at (0, 0) = 0 f(x, y) = 40x + 30y; x β₯ 0; y β₯ 0; 5x + 4y β€ 200; x + 4y β€ 120
40 roller skates and no ice skates; $1600
Sample answer: If customers cannot get ice skates, they might go somewhere else. They should combine the math model with customer needs.