Mat 501

MBA, 52nd , 01 , MAT 501 Date:04.09.
21
System of Linear Equations
Prob.-1: We plan to invest x dollars in Acme

Company bonds, which pay 6.5 percent interest ,
and y dollars in Star Company bonds, which pay 9
percent interest. If $50000 is to be invested and we
require that $4000 interest be received, how much
should be invested in each bond?
Solution: According to the question,
𝒙 + 𝒚 = 𝟓𝟎𝟎𝟎𝟎….…………(1)
𝟔.𝟓𝒙 𝟗𝒚
and + = 𝟒𝟎𝟎𝟎 ……………(2)
𝟏𝟎𝟎 𝟏𝟎𝟎
From (1) , we get
y = 50000 – x ………………(3)
From (2) , we get
𝟔. 𝟓𝒙 + 𝟗𝒚
= 𝟒𝟎𝟎𝟎
𝟏𝟎𝟎
𝑶𝒓 , 𝟔. 𝟓𝒙 + 𝟗𝒚 = 𝟒𝟎𝟎𝟎𝟎𝟎
𝑶𝒓 , 𝟔. 𝟓𝒙 = 𝟒𝟎𝟎𝟎𝟎𝟎 − 𝟗𝒚
𝑶𝒓 , 𝟔. 𝟓𝒙 = 𝟒𝟎𝟎𝟎𝟎𝟎 − 𝟗(𝟓𝟎𝟎𝟎𝟎 − 𝒙) ; by (3)
𝑶𝒓 , 𝟔. 𝟓𝒙 = 𝟒𝟎𝟎𝟎𝟎𝟎 − 𝟒𝟓𝟎𝟎𝟎𝟎 + 𝟗𝒙
𝑶𝒓 , 𝟔. 𝟓𝒙 − 𝟗𝒙 = 𝟒𝟎𝟎𝟎𝟎𝟎 − 𝟒𝟓𝟎𝟎𝟎𝟎
𝑶𝒓 , −𝟐. 𝟓𝒙 = −𝟓𝟎𝟎𝟎𝟎
𝑶𝒓 , 𝒙 = 𝟐𝟎𝟎𝟎𝟎
From (3), we get
y = 50000- 20000 = 30000
Hence $20000 should be invested in Acme company
bonds and $30000 should be invested in Star
company bonds.
Prob.-2: It takes 10 minutes to make and 20

minutes to paint one captain’s chair, whereas it
takes 5 minutes to make and 8 minutes to paint
one regular chair. If 300 minutes are available for
making these products and 500 minutes are
available for painting, how many of each chair can
be made?
Solution:
Let captain’s chair can be made x units
and regular chair can be made y units.
According to the question,
𝟏𝟎𝒙 + 𝟓𝒚 = 𝟑𝟎𝟎….……...…(1)
and 𝟐𝟎𝒙 + 𝟖𝒚 = 𝟓𝟎𝟎 ……………(2)
From[𝟐 × (𝟏)] − (𝟐), we get
𝟎 + 𝟐𝒚 = 𝟏𝟎𝟎
Or, 𝟐𝒚 = 𝟏𝟎𝟎
Or, 𝒚 = 𝟓𝟎
From (1) , we get
𝟏𝟎𝒙 + 𝟓 . 𝟓𝟎 = 𝟑𝟎𝟎
𝑶𝒓, 𝟏𝟎𝒙 = 𝟑𝟎𝟎 − 𝟐𝟓𝟎
𝑶𝒓, 𝟏𝟎𝒙 = 𝟓𝟎
𝑶𝒓, 𝒙 = 𝟓
Hence captain’s chair should be made 5 units
and regular chair should be made 50 units.
Prob.-3: Computer chips F, G, and H are each made

using 3 different kinds of transistors: A, B and C.
The number of chips to be made along with their
transistor requirements is shown in the table. How
many of each kind of chip could be made using all
available transistors?
Chip A B C
F 1 2 2
G 2 6 1
H 1 3 1
12 34 14
Solution:
Let chip F can be made x units.
chip G can be made y units.
chip H can be made z units.
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟐𝒙 + 𝟔𝒚 + 𝟑𝒛 = 𝟑𝟒
𝟐𝒙 + 𝒚 + 𝒛 = 𝟏𝟒
𝐋′𝟐 = 𝐋𝟐 + (−𝟐𝐋𝟏 )
𝐋′𝟑 = 𝐋𝟑 + (−𝟐𝐋𝟏 )
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟎 + 𝟐𝒚 + 𝒛 = 𝟏𝟎
𝟎 − 𝟑𝒚 − 𝒛 = −𝟏𝟎
𝑳′𝟑 = 𝟐𝑳𝟑 + 𝟑𝑳𝟐
𝑹𝒐𝒖𝒈𝒉
𝟎 + 𝟔𝒚 + 𝟑𝒛 = 𝟑𝟎
𝟎 − 𝟔𝒚 − 𝟐𝒛 = −𝟐𝟎
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟎 + 𝟐𝒚 + 𝒛 = 𝟏𝟎
𝟎 − 𝟎 + 𝒛 = 𝟏𝟎
This is the echelon form.
From 3rd equation, we get
𝒛 = 𝟏𝟎
From 2nd equation, we get
𝟎 + 𝟐𝒚 + 𝟏𝟎 = 𝟏𝟎
𝑶𝒓 , 𝟐𝒚 = 𝟏𝟎 − 𝟏𝟎
𝑶𝒓 , 𝟐𝒚 = 𝟎
𝑶𝒓 , 𝒚 = 𝟎
From 1st equation, we get

𝒙 + 𝟐. 𝟎 + 𝟏𝟎 = 𝟏𝟐
𝑶𝒓, 𝒙 + 𝟎 = 𝟏𝟐 − 𝟏𝟎
𝑶𝒓, 𝒙 = 𝟐
Hence chip F can be made 2 units.
chip G can not be made.
chip H can be made 10 units.
Prob.-4: Solve Prob.-3 if the requirement for
transistor B in chip G is increased 6 to 7.
Chip Number of Transistors Required per

Chip, by Type
A B C
F 1 2 2
G 2 7 1
H 1 3 1
Total 12 34 14
transistors
available
Solution:
Let chip F can be made x units.
chip G can be made y units.
chip H can be made z units.
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟐𝒙 + 𝟕𝒚 + 𝟑𝒛 = 𝟑𝟒
𝟐𝒙 + 𝒚 + 𝒛 = 𝟏𝟒
𝑳′𝟐 = 𝑳𝟐 + (−𝟐𝑳𝟏 )
𝑳′𝟑 = 𝑳𝟑 + (−𝟐𝑳𝟏 )
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟎 + 𝟑𝒚 + 𝒛 = 𝟏𝟎
𝟎 − 𝟑𝒚 − 𝒛 = −𝟏𝟎
𝑳′𝟑 = 𝑳𝟑 + 𝑳𝟐
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟎 + 𝟑𝒚 + 𝒛 = 𝟏𝟎
𝟎−𝟎+𝟎=𝟎
Or,
𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟐
𝟎 + 𝟑𝒚 + 𝒛 = 𝟏𝟎
This is the echelon form.
Here z is the free variable. Let z = a , where “a” is
any real number.
𝟎 + 𝟑𝒚 + 𝒂 = 𝟏𝟎
𝑶𝒓 , 𝟑𝒚 = 𝟏𝟎 − 𝒂
𝟏𝟎 − 𝒂
𝑶𝒓 , 𝒚 =
𝟑
𝟏𝟎 − 𝒂
𝒙 + 𝟐( ) + 𝒂 = 𝟏𝟐
𝟑
𝟐𝟎 − 𝟐𝒂
𝑶𝒓, 𝒙 + + 𝒂 = 𝟏𝟐
𝟑
𝟐𝟎 − 𝟐𝒂
𝑶𝒓, 𝒙 = 𝟏𝟐 − −𝒂
𝟑
𝟑𝟔 − 𝟐𝟎 + 𝟐𝒂 − 𝟑𝒂
𝑶𝒓, 𝒙 =
𝟑
𝟏𝟔 − 𝒂
𝑶𝒓, 𝒙 =
𝟑
We know,
𝒙≥𝟎, 𝒚≥𝟎 𝒂𝒏𝒅 𝒛≥𝟎
𝟏𝟔 − 𝒂
𝑶𝒓 , ≥𝟎
𝟑
𝑶𝒓 , 𝟏𝟔 − 𝒂 ≥ 𝟎
𝑶𝒓 , 𝟏𝟔 ≥ 𝒂
𝑶𝒓 , 𝒂 ≤ 𝟏𝟔
𝒚≥𝟎
𝟏𝟎 − 𝒂
𝑶𝒓 , ≥𝟎
𝟑
𝑶𝒓 , 𝟏𝟎 − 𝒂 ≥ 𝟎
𝑶𝒓 , 𝟏𝟎 ≥ 𝒂
𝑶𝒓 , 𝒂 ≤ 𝟏𝟎
𝒂𝒏𝒅 𝒛 ≥ 𝟎
𝑶𝒓, 𝒂 ≥ 𝟎
𝑶𝒓 , 𝟎 ≤ 𝒂
𝒂 ≤ 𝟏𝟔 , 𝒂 ≤ 𝟏𝟎 , 𝟎≤𝒂
𝟎 ≤ 𝒂 ≤ 𝟏𝟎
Prob.-5: A mixture containing x pounds of
macadamia nuts, y pounds of almonds, and z
pounds of pecans are to be made. The mixture is to
weigh 5 pounds and contain 1500 units of vitamin
and 2500 calories. The vitamin and caloric contents
of the three nuts are shown in the table.
Nut Number Units of Calories
of Pounds Vitamin per
per Pound
Pound
Macadamia x 500 300
Almonds y 200 600
Pecans z 100 700
a. Determine the number of pounds of each nut to

be used in the 5- pound mixture.
b. If the cost of nuts per pound is, respectively, $2,
$3 and $1, what is the composition and cost of
the minimum- cost mixture?
Solution:
𝒙+𝒚+𝒛=𝟓
𝟓𝟎𝟎𝒙 + 𝟐𝟎𝟎𝒚 + 𝟏𝟎𝟎𝒛 = 𝟏𝟓𝟎𝟎
𝟑𝟎𝟎𝒙 + 𝟔𝟎𝟎𝒚 + 𝟕𝟎𝟎𝒛 = 𝟐𝟓𝟎𝟎
Or ,
𝒙+𝒚+𝒛=𝟓
𝟓𝒙 + 𝟐𝒚 + 𝒛 = 𝟏𝟓
𝟑𝒙 + 𝟔𝒚 + 𝟕𝒛 = 𝟐𝟓
𝐋′𝟐 = 𝐋𝟐 + (−𝟓𝐋𝟏 )
𝐋′𝟑 = 𝐋𝟑 + (−𝟑𝐋𝟏 )
𝒙+𝒚+𝒛=𝟓
𝟎 − 𝟑𝒚 − 𝟒𝒛 = −𝟏𝟎
𝟎 + 𝟑𝒚 + 𝟒𝒛 = 𝟏𝟎
𝐋′𝟑 = 𝐋𝟑 + 𝐋𝟐
𝒙+𝒚+𝒛=𝟓
𝟎 − 𝟑𝒚 − 𝟒𝒛 = −𝟏𝟎
𝟎+𝟎+𝟎=𝟎
Or,
𝒙+𝒚+𝒛=𝟓
𝟎 − 𝟑𝒚 − 𝟒𝒛 = −𝟏𝟎
Here z is the free variable. Let z = a , where “a” is
any real number.
𝟎 − 𝟑𝒚 − 𝟒𝒂 = −𝟏𝟎
𝑶𝒓 , −𝟑𝒚 = −𝟏𝟎 + 𝟒𝒂
𝑶𝒓 , 𝟑𝒚 = 𝟏𝟎 − 𝟒𝒂
𝟏𝟎 − 𝟒𝒂
𝑶𝒓 , 𝒚 =
𝟑

𝟏𝟎 − 𝟒𝒂
𝒙+ +𝒂=𝟓
𝟑
𝟏𝟎 − 𝟒𝒂
𝑶𝒓, 𝒙 = 𝟓 − −𝒂
𝟑
𝟏𝟓 − 𝟏𝟎 + 𝟒𝒂 − 𝟑𝒂
𝑶𝒓, 𝒙 =
𝟑
𝟓+𝒂
𝑶𝒓, 𝒙 =
𝟑
We know,
𝒙≥𝟎, 𝒚≥𝟎 𝒂𝒏𝒅 𝒛≥𝟎
𝟓+𝒂
𝑶𝒓 , ≥𝟎
𝟑
𝑶𝒓 , 𝟓 + 𝒂 ≥ 𝟎
𝑶𝒓 , 𝒂 ≥ −𝟓
𝑶𝒓 , −𝟓 ≤ 𝒂
𝒚≥𝟎
𝟏𝟎 − 𝟒𝒂
𝑶𝒓 , ≥𝟎
𝟑
𝑶𝒓 , 𝟏𝟎 − 𝟒𝒂 ≥ 𝟎
𝑶𝒓 , 𝟏𝟎 ≥ 𝟒𝒂
𝑶𝒓 , 𝟒𝒂 ≤ 𝟏𝟎
𝑶𝒓 , 𝒂 ≤ 𝟐. 𝟓
𝒂𝒏𝒅 𝒛 ≥ 𝟎
𝑶𝒓, 𝒂 ≥ 𝟎
𝑶𝒓 , 𝟎 ≤ 𝒂
−𝟓 ≤ 𝒂 , 𝒂 ≤ 𝟐. 𝟓, 𝟎≤𝒂
𝟎 ≤ 𝒂 ≤ 𝟐. 𝟓
The cost function is
𝑪 = 𝟐𝒙 + 𝟑𝒚 + 𝒛
𝟓+𝒂 𝟏𝟎 − 𝟒𝒂
= 𝟐. + 𝟑. +𝒂
𝟑 𝟑
𝟏𝟎 + 𝟐𝒂 + 𝟑𝟎 − 𝟏𝟐𝒂 + 𝟑𝒂
=
𝟑
𝟒𝟎 − 𝟕𝒂
𝑪=
𝟑
Here the cost will be minimum if “a” is maximum.

Let a =2.5.
The minimum cost is

𝟒𝟎 − 𝟕 × 𝟐. 𝟓
𝑪= = $𝟕. 𝟓
𝟑
Again,
𝟓+𝒂 𝟓 + 𝟐.𝟓
𝒙= = = 2.5
𝟑 𝟑
𝟏𝟎 − 𝟒𝒂 𝟏𝟎 − 𝟒 × 𝟐. 𝟓 𝟏𝟎 − 𝟏𝟎
𝒚= = = =𝟎
𝟑 𝟑 𝟑
and z = a = 2.5
Hence the macadamia nuts should be used 2.5
pounds, Almonds should not be used and Pecans
should be used 2.5 pounds.

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MBA, 52nd , 01 , MAT 501 Date:04.09.

Prob.-1: We plan to invest x dollars in Acme

Prob.-2: It takes 10 minutes to make and 20

Prob.-3: Computer chips F, G, and H are each made

From 1st equation, we get

Chip Number of Transistors Required per

a. Determine the number of pounds of each nut to

From 1st equation, we get

Here the cost will be minimum if “a” is maximum.

The minimum cost is

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