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1 | 218-221 | 2 1
Example 1 Find the height of a given tower using mathematical modelling Solution Step 1 Given physical situation is “to find the height of a given tower” Step 2 Let AB be the given tower (Fig A |
1 | 219-222 | 1
Example 1 Find the height of a given tower using mathematical modelling Solution Step 1 Given physical situation is “to find the height of a given tower” Step 2 Let AB be the given tower (Fig A 2 |
1 | 220-223 | Solution Step 1 Given physical situation is “to find the height of a given tower” Step 2 Let AB be the given tower (Fig A 2 2) |
1 | 221-224 | Step 2 Let AB be the given tower (Fig A 2 2) Let PQ be an observer measuring the
height of the tower with his eye at P |
1 | 222-225 | 2 2) Let PQ be an observer measuring the
height of the tower with his eye at P Let PQ = h and let height of tower be H |
1 | 223-226 | 2) Let PQ be an observer measuring the
height of the tower with his eye at P Let PQ = h and let height of tower be H Let α
be the angle of elevation from the eye of the observer to the top of the tower |
1 | 224-227 | Let PQ be an observer measuring the
height of the tower with his eye at P Let PQ = h and let height of tower be H Let α
be the angle of elevation from the eye of the observer to the top of the tower Fig A |
1 | 225-228 | Let PQ = h and let height of tower be H Let α
be the angle of elevation from the eye of the observer to the top of the tower Fig A 2 |
1 | 226-229 | Let α
be the angle of elevation from the eye of the observer to the top of the tower Fig A 2 2
Rationalised 2023-24
MATHEMATICAL MODELLING 199
Let
l = PC = QB
Now
tan α = AC
H
PC
h
−l
=
or
H = h + l tan α |
1 | 227-230 | Fig A 2 2
Rationalised 2023-24
MATHEMATICAL MODELLING 199
Let
l = PC = QB
Now
tan α = AC
H
PC
h
−l
=
or
H = h + l tan α (1)
Step 3 Note that the values of the parameters h, l and α (using sextant) are known to
the observer and so (1) gives the solution of the problem |
1 | 228-231 | 2 2
Rationalised 2023-24
MATHEMATICAL MODELLING 199
Let
l = PC = QB
Now
tan α = AC
H
PC
h
−l
=
or
H = h + l tan α (1)
Step 3 Note that the values of the parameters h, l and α (using sextant) are known to
the observer and so (1) gives the solution of the problem Step 4 In case, if the foot of the tower is not accessible, i |
1 | 229-232 | 2
Rationalised 2023-24
MATHEMATICAL MODELLING 199
Let
l = PC = QB
Now
tan α = AC
H
PC
h
−l
=
or
H = h + l tan α (1)
Step 3 Note that the values of the parameters h, l and α (using sextant) are known to
the observer and so (1) gives the solution of the problem Step 4 In case, if the foot of the tower is not accessible, i e |
1 | 230-233 | (1)
Step 3 Note that the values of the parameters h, l and α (using sextant) are known to
the observer and so (1) gives the solution of the problem Step 4 In case, if the foot of the tower is not accessible, i e , when l is not known to the
observer, let β be the angle of depression from P to the foot B of the tower |
1 | 231-234 | Step 4 In case, if the foot of the tower is not accessible, i e , when l is not known to the
observer, let β be the angle of depression from P to the foot B of the tower So from
∆PQB, we have
PQ
tan
QB
lh
β =
=
or l = h cot β
Step 5 is not required in this situation as exact values of the parameters h, l, α and β
are known |
1 | 232-235 | e , when l is not known to the
observer, let β be the angle of depression from P to the foot B of the tower So from
∆PQB, we have
PQ
tan
QB
lh
β =
=
or l = h cot β
Step 5 is not required in this situation as exact values of the parameters h, l, α and β
are known Example 2 Let a business firm produces three types of products P1, P2 and P3 that
uses three types of raw materials R1, R2 and R3 |
1 | 233-236 | , when l is not known to the
observer, let β be the angle of depression from P to the foot B of the tower So from
∆PQB, we have
PQ
tan
QB
lh
β =
=
or l = h cot β
Step 5 is not required in this situation as exact values of the parameters h, l, α and β
are known Example 2 Let a business firm produces three types of products P1, P2 and P3 that
uses three types of raw materials R1, R2 and R3 Let the firm has purchase orders from
two clients F1 and F2 |
1 | 234-237 | So from
∆PQB, we have
PQ
tan
QB
lh
β =
=
or l = h cot β
Step 5 is not required in this situation as exact values of the parameters h, l, α and β
are known Example 2 Let a business firm produces three types of products P1, P2 and P3 that
uses three types of raw materials R1, R2 and R3 Let the firm has purchase orders from
two clients F1 and F2 Considering the situation that the firm has a limited quantity of
R1, R2 and R3, respectively, prepare a model to determine the quantities of the raw
material R1, R2 and R3 required to meet the purchase orders |
1 | 235-238 | Example 2 Let a business firm produces three types of products P1, P2 and P3 that
uses three types of raw materials R1, R2 and R3 Let the firm has purchase orders from
two clients F1 and F2 Considering the situation that the firm has a limited quantity of
R1, R2 and R3, respectively, prepare a model to determine the quantities of the raw
material R1, R2 and R3 required to meet the purchase orders Solution Step 1 The physical situation is well identified in the problem |
1 | 236-239 | Let the firm has purchase orders from
two clients F1 and F2 Considering the situation that the firm has a limited quantity of
R1, R2 and R3, respectively, prepare a model to determine the quantities of the raw
material R1, R2 and R3 required to meet the purchase orders Solution Step 1 The physical situation is well identified in the problem Step 2 Let A be a matrix that represents purchase orders from the two clients F1 and
F2 |
1 | 237-240 | Considering the situation that the firm has a limited quantity of
R1, R2 and R3, respectively, prepare a model to determine the quantities of the raw
material R1, R2 and R3 required to meet the purchase orders Solution Step 1 The physical situation is well identified in the problem Step 2 Let A be a matrix that represents purchase orders from the two clients F1 and
F2 Then, A is of the form
1
2
3
1
2
P P
P
F •
•
•
A
F
•
•
•
=
Let B be the matrix that represents the amount of raw materials R1, R2 and R3,
required to manufacture each unit of the products P1, P2 and P3 |
1 | 238-241 | Solution Step 1 The physical situation is well identified in the problem Step 2 Let A be a matrix that represents purchase orders from the two clients F1 and
F2 Then, A is of the form
1
2
3
1
2
P P
P
F •
•
•
A
F
•
•
•
=
Let B be the matrix that represents the amount of raw materials R1, R2 and R3,
required to manufacture each unit of the products P1, P2 and P3 Then, B is of the form
1
2
3
1
2
3
•R R R
•
•
P
B
P
•
•
•
P
•
•
•
=
Rationalised 2023-24
200
MATHEMATICS
Step 3 Note that the product (which in this case is well defined) of matrices A and B
is given by the following matrix
1
2
3
1
2
R R R
F •
•
•
AB F
•
•
•
=
which in fact gives the desired quantities of the raw materials R1, R2 and R3 to fulfill
the purchase orders of the two clients F1 and F2 |
1 | 239-242 | Step 2 Let A be a matrix that represents purchase orders from the two clients F1 and
F2 Then, A is of the form
1
2
3
1
2
P P
P
F •
•
•
A
F
•
•
•
=
Let B be the matrix that represents the amount of raw materials R1, R2 and R3,
required to manufacture each unit of the products P1, P2 and P3 Then, B is of the form
1
2
3
1
2
3
•R R R
•
•
P
B
P
•
•
•
P
•
•
•
=
Rationalised 2023-24
200
MATHEMATICS
Step 3 Note that the product (which in this case is well defined) of matrices A and B
is given by the following matrix
1
2
3
1
2
R R R
F •
•
•
AB F
•
•
•
=
which in fact gives the desired quantities of the raw materials R1, R2 and R3 to fulfill
the purchase orders of the two clients F1 and F2 Example 3 Interpret the model in Example 2, in case
3
4
0
10
15
6
A =
, B
7
9
3
10
20
0
5
12
7
=
and the available raw materials are 330 units of R1, 455 units of R2 and 140 units of R3 |
1 | 240-243 | Then, A is of the form
1
2
3
1
2
P P
P
F •
•
•
A
F
•
•
•
=
Let B be the matrix that represents the amount of raw materials R1, R2 and R3,
required to manufacture each unit of the products P1, P2 and P3 Then, B is of the form
1
2
3
1
2
3
•R R R
•
•
P
B
P
•
•
•
P
•
•
•
=
Rationalised 2023-24
200
MATHEMATICS
Step 3 Note that the product (which in this case is well defined) of matrices A and B
is given by the following matrix
1
2
3
1
2
R R R
F •
•
•
AB F
•
•
•
=
which in fact gives the desired quantities of the raw materials R1, R2 and R3 to fulfill
the purchase orders of the two clients F1 and F2 Example 3 Interpret the model in Example 2, in case
3
4
0
10
15
6
A =
, B
7
9
3
10
20
0
5
12
7
=
and the available raw materials are 330 units of R1, 455 units of R2 and 140 units of R3 Solution Note that
AB =
3
4
0
10
15
6
7
9
3
10
20
0
5
12
7
=
R
R
R
F
F
1
2
3
1
2
165
247
87
170
220
60
This clearly shows that to meet the purchase order of F1 and F2, the raw material
required is 335 units of R1, 467 units of R2 and 147 units of R3 which is much more than
the available raw material |
1 | 241-244 | Then, B is of the form
1
2
3
1
2
3
•R R R
•
•
P
B
P
•
•
•
P
•
•
•
=
Rationalised 2023-24
200
MATHEMATICS
Step 3 Note that the product (which in this case is well defined) of matrices A and B
is given by the following matrix
1
2
3
1
2
R R R
F •
•
•
AB F
•
•
•
=
which in fact gives the desired quantities of the raw materials R1, R2 and R3 to fulfill
the purchase orders of the two clients F1 and F2 Example 3 Interpret the model in Example 2, in case
3
4
0
10
15
6
A =
, B
7
9
3
10
20
0
5
12
7
=
and the available raw materials are 330 units of R1, 455 units of R2 and 140 units of R3 Solution Note that
AB =
3
4
0
10
15
6
7
9
3
10
20
0
5
12
7
=
R
R
R
F
F
1
2
3
1
2
165
247
87
170
220
60
This clearly shows that to meet the purchase order of F1 and F2, the raw material
required is 335 units of R1, 467 units of R2 and 147 units of R3 which is much more than
the available raw material Since the amount of raw material required to manufacture
each unit of the three products is fixed, we can either ask for an increase in the
available raw material or we may ask the clients to reduce their orders |
1 | 242-245 | Example 3 Interpret the model in Example 2, in case
3
4
0
10
15
6
A =
, B
7
9
3
10
20
0
5
12
7
=
and the available raw materials are 330 units of R1, 455 units of R2 and 140 units of R3 Solution Note that
AB =
3
4
0
10
15
6
7
9
3
10
20
0
5
12
7
=
R
R
R
F
F
1
2
3
1
2
165
247
87
170
220
60
This clearly shows that to meet the purchase order of F1 and F2, the raw material
required is 335 units of R1, 467 units of R2 and 147 units of R3 which is much more than
the available raw material Since the amount of raw material required to manufacture
each unit of the three products is fixed, we can either ask for an increase in the
available raw material or we may ask the clients to reduce their orders Remark If we replace A in Example 3 by A1 given by
A1 = 9
12
6
10
20
0
i |
1 | 243-246 | Solution Note that
AB =
3
4
0
10
15
6
7
9
3
10
20
0
5
12
7
=
R
R
R
F
F
1
2
3
1
2
165
247
87
170
220
60
This clearly shows that to meet the purchase order of F1 and F2, the raw material
required is 335 units of R1, 467 units of R2 and 147 units of R3 which is much more than
the available raw material Since the amount of raw material required to manufacture
each unit of the three products is fixed, we can either ask for an increase in the
available raw material or we may ask the clients to reduce their orders Remark If we replace A in Example 3 by A1 given by
A1 = 9
12
6
10
20
0
i e |
1 | 244-247 | Since the amount of raw material required to manufacture
each unit of the three products is fixed, we can either ask for an increase in the
available raw material or we may ask the clients to reduce their orders Remark If we replace A in Example 3 by A1 given by
A1 = 9
12
6
10
20
0
i e , if the clients agree to reduce their purchase orders, then
A1 B =
3
4
0
9
12
6
7
9
3
10
20
0
5
12
7
=
141
216
78
170
220
60
Rationalised 2023-24
MATHEMATICAL MODELLING 201
This requires 311 units of R1, 436 units of R2 and 138 units of R3 which are well
below the available raw materials, i |
1 | 245-248 | Remark If we replace A in Example 3 by A1 given by
A1 = 9
12
6
10
20
0
i e , if the clients agree to reduce their purchase orders, then
A1 B =
3
4
0
9
12
6
7
9
3
10
20
0
5
12
7
=
141
216
78
170
220
60
Rationalised 2023-24
MATHEMATICAL MODELLING 201
This requires 311 units of R1, 436 units of R2 and 138 units of R3 which are well
below the available raw materials, i e |
1 | 246-249 | e , if the clients agree to reduce their purchase orders, then
A1 B =
3
4
0
9
12
6
7
9
3
10
20
0
5
12
7
=
141
216
78
170
220
60
Rationalised 2023-24
MATHEMATICAL MODELLING 201
This requires 311 units of R1, 436 units of R2 and 138 units of R3 which are well
below the available raw materials, i e , 330 units of R1, 455 units of R2 and 140 units of
R3 |
1 | 247-250 | , if the clients agree to reduce their purchase orders, then
A1 B =
3
4
0
9
12
6
7
9
3
10
20
0
5
12
7
=
141
216
78
170
220
60
Rationalised 2023-24
MATHEMATICAL MODELLING 201
This requires 311 units of R1, 436 units of R2 and 138 units of R3 which are well
below the available raw materials, i e , 330 units of R1, 455 units of R2 and 140 units of
R3 Thus, if the revised purchase orders of the clients are given by A1, then the firm
can easily supply the purchase orders of the two clients |
1 | 248-251 | e , 330 units of R1, 455 units of R2 and 140 units of
R3 Thus, if the revised purchase orders of the clients are given by A1, then the firm
can easily supply the purchase orders of the two clients ANote One may further modify A so as to make full use of the available
raw material |
1 | 249-252 | , 330 units of R1, 455 units of R2 and 140 units of
R3 Thus, if the revised purchase orders of the clients are given by A1, then the firm
can easily supply the purchase orders of the two clients ANote One may further modify A so as to make full use of the available
raw material Query Can we make a mathematical model with a given B and with fixed quantities of
the available raw material that can help the firm owner to ask the clients to modify their
orders in such a way that the firm makes the full use of its available raw material |
1 | 250-253 | Thus, if the revised purchase orders of the clients are given by A1, then the firm
can easily supply the purchase orders of the two clients ANote One may further modify A so as to make full use of the available
raw material Query Can we make a mathematical model with a given B and with fixed quantities of
the available raw material that can help the firm owner to ask the clients to modify their
orders in such a way that the firm makes the full use of its available raw material The answer to this query is given in the following example:
Example 4 Suppose P1, P2, P3 and R1, R2, R3 are as in Example 2 |
1 | 251-254 | ANote One may further modify A so as to make full use of the available
raw material Query Can we make a mathematical model with a given B and with fixed quantities of
the available raw material that can help the firm owner to ask the clients to modify their
orders in such a way that the firm makes the full use of its available raw material The answer to this query is given in the following example:
Example 4 Suppose P1, P2, P3 and R1, R2, R3 are as in Example 2 Let the firm has
330 units of R1, 455 units of R2 and 140 units of R3 available with it and let the amount
of raw materials R1, R2 and R3 required to manufacture each unit of the three products
is given by
1
2
3
1
2
3
R
R
R
3
4
0
P
B
P
7
9
3
P
5
12
7
=
How many units of each product is to be made so as to utilise the full available raw
material |
1 | 252-255 | Query Can we make a mathematical model with a given B and with fixed quantities of
the available raw material that can help the firm owner to ask the clients to modify their
orders in such a way that the firm makes the full use of its available raw material The answer to this query is given in the following example:
Example 4 Suppose P1, P2, P3 and R1, R2, R3 are as in Example 2 Let the firm has
330 units of R1, 455 units of R2 and 140 units of R3 available with it and let the amount
of raw materials R1, R2 and R3 required to manufacture each unit of the three products
is given by
1
2
3
1
2
3
R
R
R
3
4
0
P
B
P
7
9
3
P
5
12
7
=
How many units of each product is to be made so as to utilise the full available raw
material Solution Step 1 The situation is easily identifiable |
1 | 253-256 | The answer to this query is given in the following example:
Example 4 Suppose P1, P2, P3 and R1, R2, R3 are as in Example 2 Let the firm has
330 units of R1, 455 units of R2 and 140 units of R3 available with it and let the amount
of raw materials R1, R2 and R3 required to manufacture each unit of the three products
is given by
1
2
3
1
2
3
R
R
R
3
4
0
P
B
P
7
9
3
P
5
12
7
=
How many units of each product is to be made so as to utilise the full available raw
material Solution Step 1 The situation is easily identifiable Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3 |
1 | 254-257 | Let the firm has
330 units of R1, 455 units of R2 and 140 units of R3 available with it and let the amount
of raw materials R1, R2 and R3 required to manufacture each unit of the three products
is given by
1
2
3
1
2
3
R
R
R
3
4
0
P
B
P
7
9
3
P
5
12
7
=
How many units of each product is to be made so as to utilise the full available raw
material Solution Step 1 The situation is easily identifiable Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3 Since
product P1 requires 3 units of R1, P2 requires 7 units of R1 and P3 requires 5 units of R1
(observe matrix B) and the total number of units, of R1, available is 330, we have
3x + 7y + 5z = 330 (for raw material R1)
Similarly, we have
4x + 9y + 12z = 455 (for raw material R2)
and
3y + 7z = 140 (for raw material R3)
This system of equations can be expressed in matrix form as
3
7
5
4
9
12
0
3
7
330
455
140
=
x
y
z
Rationalised 2023-24
202
MATHEMATICS
Step 3 Using elementary row operations, we obtain
1
0
0
0
1
0
0
0
1
20
35
5
=
x
y
z
This gives x = 20, y = 35 and z = 5 |
1 | 255-258 | Solution Step 1 The situation is easily identifiable Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3 Since
product P1 requires 3 units of R1, P2 requires 7 units of R1 and P3 requires 5 units of R1
(observe matrix B) and the total number of units, of R1, available is 330, we have
3x + 7y + 5z = 330 (for raw material R1)
Similarly, we have
4x + 9y + 12z = 455 (for raw material R2)
and
3y + 7z = 140 (for raw material R3)
This system of equations can be expressed in matrix form as
3
7
5
4
9
12
0
3
7
330
455
140
=
x
y
z
Rationalised 2023-24
202
MATHEMATICS
Step 3 Using elementary row operations, we obtain
1
0
0
0
1
0
0
0
1
20
35
5
=
x
y
z
This gives x = 20, y = 35 and z = 5 Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material |
1 | 256-259 | Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3 Since
product P1 requires 3 units of R1, P2 requires 7 units of R1 and P3 requires 5 units of R1
(observe matrix B) and the total number of units, of R1, available is 330, we have
3x + 7y + 5z = 330 (for raw material R1)
Similarly, we have
4x + 9y + 12z = 455 (for raw material R2)
and
3y + 7z = 140 (for raw material R3)
This system of equations can be expressed in matrix form as
3
7
5
4
9
12
0
3
7
330
455
140
=
x
y
z
Rationalised 2023-24
202
MATHEMATICS
Step 3 Using elementary row operations, we obtain
1
0
0
0
1
0
0
0
1
20
35
5
=
x
y
z
This gives x = 20, y = 35 and z = 5 Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material Remark One may observe that if the manufacturer decides to manufacture according
to the available raw material and not according to the purchase orders of the two
clients F1 and F2 (as in Example 3), he/she is unable to meet these purchase orders as
F1 demanded 6 units of P3 where as the manufacturer can make only 5 units of P3 |
1 | 257-260 | Since
product P1 requires 3 units of R1, P2 requires 7 units of R1 and P3 requires 5 units of R1
(observe matrix B) and the total number of units, of R1, available is 330, we have
3x + 7y + 5z = 330 (for raw material R1)
Similarly, we have
4x + 9y + 12z = 455 (for raw material R2)
and
3y + 7z = 140 (for raw material R3)
This system of equations can be expressed in matrix form as
3
7
5
4
9
12
0
3
7
330
455
140
=
x
y
z
Rationalised 2023-24
202
MATHEMATICS
Step 3 Using elementary row operations, we obtain
1
0
0
0
1
0
0
0
1
20
35
5
=
x
y
z
This gives x = 20, y = 35 and z = 5 Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material Remark One may observe that if the manufacturer decides to manufacture according
to the available raw material and not according to the purchase orders of the two
clients F1 and F2 (as in Example 3), he/she is unable to meet these purchase orders as
F1 demanded 6 units of P3 where as the manufacturer can make only 5 units of P3 Example 5 A manufacturer of medicines is preparing a production plan of medicines
M1 and M2 |
1 | 258-261 | Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material Remark One may observe that if the manufacturer decides to manufacture according
to the available raw material and not according to the purchase orders of the two
clients F1 and F2 (as in Example 3), he/she is unable to meet these purchase orders as
F1 demanded 6 units of P3 where as the manufacturer can make only 5 units of P3 Example 5 A manufacturer of medicines is preparing a production plan of medicines
M1 and M2 There are sufficient raw materials available to make 20000 bottles of M1
and 40000 bottles of M2, but there are only 45000 bottles into which either of the
medicines can be put |
1 | 259-262 | Remark One may observe that if the manufacturer decides to manufacture according
to the available raw material and not according to the purchase orders of the two
clients F1 and F2 (as in Example 3), he/she is unable to meet these purchase orders as
F1 demanded 6 units of P3 where as the manufacturer can make only 5 units of P3 Example 5 A manufacturer of medicines is preparing a production plan of medicines
M1 and M2 There are sufficient raw materials available to make 20000 bottles of M1
and 40000 bottles of M2, but there are only 45000 bottles into which either of the
medicines can be put Further, it takes 3 hours to prepare enough material to fill 1000
bottles of M1, it takes 1 hour to prepare enough material to fill 1000 bottles of M2 and
there are 66 hours available for this operation |
1 | 260-263 | Example 5 A manufacturer of medicines is preparing a production plan of medicines
M1 and M2 There are sufficient raw materials available to make 20000 bottles of M1
and 40000 bottles of M2, but there are only 45000 bottles into which either of the
medicines can be put Further, it takes 3 hours to prepare enough material to fill 1000
bottles of M1, it takes 1 hour to prepare enough material to fill 1000 bottles of M2 and
there are 66 hours available for this operation The profit is Rs 8 per bottle for M1 and
Rs 7 per bottle for M2 |
1 | 261-264 | There are sufficient raw materials available to make 20000 bottles of M1
and 40000 bottles of M2, but there are only 45000 bottles into which either of the
medicines can be put Further, it takes 3 hours to prepare enough material to fill 1000
bottles of M1, it takes 1 hour to prepare enough material to fill 1000 bottles of M2 and
there are 66 hours available for this operation The profit is Rs 8 per bottle for M1 and
Rs 7 per bottle for M2 How should the manufacturer schedule his/her production in
order to maximise profit |
1 | 262-265 | Further, it takes 3 hours to prepare enough material to fill 1000
bottles of M1, it takes 1 hour to prepare enough material to fill 1000 bottles of M2 and
there are 66 hours available for this operation The profit is Rs 8 per bottle for M1 and
Rs 7 per bottle for M2 How should the manufacturer schedule his/her production in
order to maximise profit Solution Step 1 To find the number of bottles of M1 and M2 in order to maximise the
profit under the given hypotheses |
1 | 263-266 | The profit is Rs 8 per bottle for M1 and
Rs 7 per bottle for M2 How should the manufacturer schedule his/her production in
order to maximise profit Solution Step 1 To find the number of bottles of M1 and M2 in order to maximise the
profit under the given hypotheses Step 2 Let x be the number of bottles of type M1 medicine and y be the number of
bottles of type M2 medicine |
1 | 264-267 | How should the manufacturer schedule his/her production in
order to maximise profit Solution Step 1 To find the number of bottles of M1 and M2 in order to maximise the
profit under the given hypotheses Step 2 Let x be the number of bottles of type M1 medicine and y be the number of
bottles of type M2 medicine Since profit is Rs 8 per bottle for M1 and Rs 7 per bottle
for M2, therefore the objective function (which is to be maximised) is given by
Z ≡ Z (x, y) = 8x + 7y
The objective function is to be maximised subject to the constraints (Refer Chapter
12 on Linear Programming)
x
y
x
xy
y
x
y
≤
≤
+
+≤
≤
≥
≥
20000
40000
45000
3
66000
0
0
, |
1 | 265-268 | Solution Step 1 To find the number of bottles of M1 and M2 in order to maximise the
profit under the given hypotheses Step 2 Let x be the number of bottles of type M1 medicine and y be the number of
bottles of type M2 medicine Since profit is Rs 8 per bottle for M1 and Rs 7 per bottle
for M2, therefore the objective function (which is to be maximised) is given by
Z ≡ Z (x, y) = 8x + 7y
The objective function is to be maximised subject to the constraints (Refer Chapter
12 on Linear Programming)
x
y
x
xy
y
x
y
≤
≤
+
+≤
≤
≥
≥
20000
40000
45000
3
66000
0
0
, (1)
Step 3 The shaded region OPQRST is the feasible region for the constraints (1)
(Fig A |
1 | 266-269 | Step 2 Let x be the number of bottles of type M1 medicine and y be the number of
bottles of type M2 medicine Since profit is Rs 8 per bottle for M1 and Rs 7 per bottle
for M2, therefore the objective function (which is to be maximised) is given by
Z ≡ Z (x, y) = 8x + 7y
The objective function is to be maximised subject to the constraints (Refer Chapter
12 on Linear Programming)
x
y
x
xy
y
x
y
≤
≤
+
+≤
≤
≥
≥
20000
40000
45000
3
66000
0
0
, (1)
Step 3 The shaded region OPQRST is the feasible region for the constraints (1)
(Fig A 2 |
1 | 267-270 | Since profit is Rs 8 per bottle for M1 and Rs 7 per bottle
for M2, therefore the objective function (which is to be maximised) is given by
Z ≡ Z (x, y) = 8x + 7y
The objective function is to be maximised subject to the constraints (Refer Chapter
12 on Linear Programming)
x
y
x
xy
y
x
y
≤
≤
+
+≤
≤
≥
≥
20000
40000
45000
3
66000
0
0
, (1)
Step 3 The shaded region OPQRST is the feasible region for the constraints (1)
(Fig A 2 3) |
1 | 268-271 | (1)
Step 3 The shaded region OPQRST is the feasible region for the constraints (1)
(Fig A 2 3) The co-ordinates of vertices O, P, Q, R, S and T are (0, 0), (20000, 0),
(20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respectively |
1 | 269-272 | 2 3) The co-ordinates of vertices O, P, Q, R, S and T are (0, 0), (20000, 0),
(20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respectively Rationalised 2023-24
MATHEMATICAL MODELLING 203
Fig A |
1 | 270-273 | 3) The co-ordinates of vertices O, P, Q, R, S and T are (0, 0), (20000, 0),
(20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respectively Rationalised 2023-24
MATHEMATICAL MODELLING 203
Fig A 2 |
1 | 271-274 | The co-ordinates of vertices O, P, Q, R, S and T are (0, 0), (20000, 0),
(20000, 6000), (10500, 34500), (5000, 40000) and (0, 40000), respectively Rationalised 2023-24
MATHEMATICAL MODELLING 203
Fig A 2 3
Note that
Z at P (0, 0) = 0
Z at P (20000, 0) = 8 × 20000 = 160000
Z at Q (20000, 6000) = 8 × 20000 + 7 × 6000 = 202000
Z at R (10500, 34500) = 8 × 10500 + 7 × 34500 = 325500
Z at S = (5000, 40000) = 8 × 5000 + 7 × 40000 = 320000
Z at T = (0, 40000) = 7 × 40000 = 280000
Now observe that the profit is maximum at x = 10500 and y = 34500 and the
maximum profit is ` 325500 |
1 | 272-275 | Rationalised 2023-24
MATHEMATICAL MODELLING 203
Fig A 2 3
Note that
Z at P (0, 0) = 0
Z at P (20000, 0) = 8 × 20000 = 160000
Z at Q (20000, 6000) = 8 × 20000 + 7 × 6000 = 202000
Z at R (10500, 34500) = 8 × 10500 + 7 × 34500 = 325500
Z at S = (5000, 40000) = 8 × 5000 + 7 × 40000 = 320000
Z at T = (0, 40000) = 7 × 40000 = 280000
Now observe that the profit is maximum at x = 10500 and y = 34500 and the
maximum profit is ` 325500 Hence, the manufacturer should produce 10500 bottles of
M1 medicine and 34500 bottles of M2 medicine in order to get maximum profit of
` 325500 |
1 | 273-276 | 2 3
Note that
Z at P (0, 0) = 0
Z at P (20000, 0) = 8 × 20000 = 160000
Z at Q (20000, 6000) = 8 × 20000 + 7 × 6000 = 202000
Z at R (10500, 34500) = 8 × 10500 + 7 × 34500 = 325500
Z at S = (5000, 40000) = 8 × 5000 + 7 × 40000 = 320000
Z at T = (0, 40000) = 7 × 40000 = 280000
Now observe that the profit is maximum at x = 10500 and y = 34500 and the
maximum profit is ` 325500 Hence, the manufacturer should produce 10500 bottles of
M1 medicine and 34500 bottles of M2 medicine in order to get maximum profit of
` 325500 Example 6 Suppose a company plans to produce a new product that incur some costs
(fixed and variable) and let the company plans to sell the product at a fixed price |
1 | 274-277 | 3
Note that
Z at P (0, 0) = 0
Z at P (20000, 0) = 8 × 20000 = 160000
Z at Q (20000, 6000) = 8 × 20000 + 7 × 6000 = 202000
Z at R (10500, 34500) = 8 × 10500 + 7 × 34500 = 325500
Z at S = (5000, 40000) = 8 × 5000 + 7 × 40000 = 320000
Z at T = (0, 40000) = 7 × 40000 = 280000
Now observe that the profit is maximum at x = 10500 and y = 34500 and the
maximum profit is ` 325500 Hence, the manufacturer should produce 10500 bottles of
M1 medicine and 34500 bottles of M2 medicine in order to get maximum profit of
` 325500 Example 6 Suppose a company plans to produce a new product that incur some costs
(fixed and variable) and let the company plans to sell the product at a fixed price Prepare a mathematical model to examine the profitability |
1 | 275-278 | Hence, the manufacturer should produce 10500 bottles of
M1 medicine and 34500 bottles of M2 medicine in order to get maximum profit of
` 325500 Example 6 Suppose a company plans to produce a new product that incur some costs
(fixed and variable) and let the company plans to sell the product at a fixed price Prepare a mathematical model to examine the profitability Solution Step 1 Situation is clearly identifiable |
1 | 276-279 | Example 6 Suppose a company plans to produce a new product that incur some costs
(fixed and variable) and let the company plans to sell the product at a fixed price Prepare a mathematical model to examine the profitability Solution Step 1 Situation is clearly identifiable Rationalised 2023-24
204
MATHEMATICS
Step 2 Formulation: We are given that the costs are of two types: fixed and variable |
1 | 277-280 | Prepare a mathematical model to examine the profitability Solution Step 1 Situation is clearly identifiable Rationalised 2023-24
204
MATHEMATICS
Step 2 Formulation: We are given that the costs are of two types: fixed and variable The fixed costs are independent of the number of units produced (e |
1 | 278-281 | Solution Step 1 Situation is clearly identifiable Rationalised 2023-24
204
MATHEMATICS
Step 2 Formulation: We are given that the costs are of two types: fixed and variable The fixed costs are independent of the number of units produced (e g |
1 | 279-282 | Rationalised 2023-24
204
MATHEMATICS
Step 2 Formulation: We are given that the costs are of two types: fixed and variable The fixed costs are independent of the number of units produced (e g , rent and rates),
while the variable costs increase with the number of units produced (e |
1 | 280-283 | The fixed costs are independent of the number of units produced (e g , rent and rates),
while the variable costs increase with the number of units produced (e g |
1 | 281-284 | g , rent and rates),
while the variable costs increase with the number of units produced (e g , material) |
1 | 282-285 | , rent and rates),
while the variable costs increase with the number of units produced (e g , material) Initially, we assume that the variable costs are directly proportional to the number of
units produced — this should simplify our model |
1 | 283-286 | g , material) Initially, we assume that the variable costs are directly proportional to the number of
units produced — this should simplify our model The company earn a certain amount
of money by selling its products and wants to ensure that it is maximum |
1 | 284-287 | , material) Initially, we assume that the variable costs are directly proportional to the number of
units produced — this should simplify our model The company earn a certain amount
of money by selling its products and wants to ensure that it is maximum For convenience,
we assume that all units produced are sold immediately |
1 | 285-288 | Initially, we assume that the variable costs are directly proportional to the number of
units produced — this should simplify our model The company earn a certain amount
of money by selling its products and wants to ensure that it is maximum For convenience,
we assume that all units produced are sold immediately The mathematical model
Let
x = number of units produced and sold
C = total cost of production (in rupees)
I = income from sales (in rupees)
P = profit (in rupees)
Our assumptions above state that C consists of two parts:
(i)
fixed cost = a (in rupees),
(ii)
Thenvariable cost = b (rupees/unit produced) |
1 | 286-289 | The company earn a certain amount
of money by selling its products and wants to ensure that it is maximum For convenience,
we assume that all units produced are sold immediately The mathematical model
Let
x = number of units produced and sold
C = total cost of production (in rupees)
I = income from sales (in rupees)
P = profit (in rupees)
Our assumptions above state that C consists of two parts:
(i)
fixed cost = a (in rupees),
(ii)
Thenvariable cost = b (rupees/unit produced) C = a + bx |
1 | 287-290 | For convenience,
we assume that all units produced are sold immediately The mathematical model
Let
x = number of units produced and sold
C = total cost of production (in rupees)
I = income from sales (in rupees)
P = profit (in rupees)
Our assumptions above state that C consists of two parts:
(i)
fixed cost = a (in rupees),
(ii)
Thenvariable cost = b (rupees/unit produced) C = a + bx (1)
Also, income I depends on selling price s (rupees/unit)
Thus
I = sx |
1 | 288-291 | The mathematical model
Let
x = number of units produced and sold
C = total cost of production (in rupees)
I = income from sales (in rupees)
P = profit (in rupees)
Our assumptions above state that C consists of two parts:
(i)
fixed cost = a (in rupees),
(ii)
Thenvariable cost = b (rupees/unit produced) C = a + bx (1)
Also, income I depends on selling price s (rupees/unit)
Thus
I = sx (2)
The profit P is then the difference between income and costs |
1 | 289-292 | C = a + bx (1)
Also, income I depends on selling price s (rupees/unit)
Thus
I = sx (2)
The profit P is then the difference between income and costs So
P = I – C
= sx – (a + bx)
= (s – b) x – a |
1 | 290-293 | (1)
Also, income I depends on selling price s (rupees/unit)
Thus
I = sx (2)
The profit P is then the difference between income and costs So
P = I – C
= sx – (a + bx)
= (s – b) x – a (3)
We now have a mathematical model of the relationships (1) to (3) between
the variables x, C, I, P, a, b, s |
1 | 291-294 | (2)
The profit P is then the difference between income and costs So
P = I – C
= sx – (a + bx)
= (s – b) x – a (3)
We now have a mathematical model of the relationships (1) to (3) between
the variables x, C, I, P, a, b, s These variables may be classified as:
independent
x
dependent
C, I, P
parameters
a, b, s
The manufacturer, knowing x, a, b, s can determine P |
1 | 292-295 | So
P = I – C
= sx – (a + bx)
= (s – b) x – a (3)
We now have a mathematical model of the relationships (1) to (3) between
the variables x, C, I, P, a, b, s These variables may be classified as:
independent
x
dependent
C, I, P
parameters
a, b, s
The manufacturer, knowing x, a, b, s can determine P Step 3 From (3), we can observe that for the break even point (i |
1 | 293-296 | (3)
We now have a mathematical model of the relationships (1) to (3) between
the variables x, C, I, P, a, b, s These variables may be classified as:
independent
x
dependent
C, I, P
parameters
a, b, s
The manufacturer, knowing x, a, b, s can determine P Step 3 From (3), we can observe that for the break even point (i e |
1 | 294-297 | These variables may be classified as:
independent
x
dependent
C, I, P
parameters
a, b, s
The manufacturer, knowing x, a, b, s can determine P Step 3 From (3), we can observe that for the break even point (i e , make neither profit
nor loss), he must have P = 0, i |
1 | 295-298 | Step 3 From (3), we can observe that for the break even point (i e , make neither profit
nor loss), he must have P = 0, i e |
1 | 296-299 | e , make neither profit
nor loss), he must have P = 0, i e ,
units |
1 | 297-300 | , make neither profit
nor loss), he must have P = 0, i e ,
units a
x
s
b
= −
Steps 4 and 5 In view of the break even point, one may conclude that if the company
produces few units, i |
1 | 298-301 | e ,
units a
x
s
b
= −
Steps 4 and 5 In view of the break even point, one may conclude that if the company
produces few units, i e |
1 | 299-302 | ,
units a
x
s
b
= −
Steps 4 and 5 In view of the break even point, one may conclude that if the company
produces few units, i e , less than
units
a
x
s b
= −
, then the company will suffer loss
Rationalised 2023-24
MATHEMATICAL MODELLING 205
and if it produces large number of units, i |
1 | 300-303 | a
x
s
b
= −
Steps 4 and 5 In view of the break even point, one may conclude that if the company
produces few units, i e , less than
units
a
x
s b
= −
, then the company will suffer loss
Rationalised 2023-24
MATHEMATICAL MODELLING 205
and if it produces large number of units, i e |
1 | 301-304 | e , less than
units
a
x
s b
= −
, then the company will suffer loss
Rationalised 2023-24
MATHEMATICAL MODELLING 205
and if it produces large number of units, i e , much more than
units
a
s b
−
, then it can
make huge profit |
1 | 302-305 | , less than
units
a
x
s b
= −
, then the company will suffer loss
Rationalised 2023-24
MATHEMATICAL MODELLING 205
and if it produces large number of units, i e , much more than
units
a
s b
−
, then it can
make huge profit Further, if the break even point proves to be unrealistic, then another
model could be tried or the assumptions regarding cash flow may be modified |
1 | 303-306 | e , much more than
units
a
s b
−
, then it can
make huge profit Further, if the break even point proves to be unrealistic, then another
model could be tried or the assumptions regarding cash flow may be modified Remark From (3), we also have
dP
s
b
d x =
−
This means that rate of change of P with respect to x depends on the quantity
s – b, which is the difference of selling price and the variable cost of each product |
1 | 304-307 | , much more than
units
a
s b
−
, then it can
make huge profit Further, if the break even point proves to be unrealistic, then another
model could be tried or the assumptions regarding cash flow may be modified Remark From (3), we also have
dP
s
b
d x =
−
This means that rate of change of P with respect to x depends on the quantity
s – b, which is the difference of selling price and the variable cost of each product Thus, in order to gain profit, this should be positive and to get large gains, we need to
produce large quantity of the product and at the same time try to reduce the variable
cost |
1 | 305-308 | Further, if the break even point proves to be unrealistic, then another
model could be tried or the assumptions regarding cash flow may be modified Remark From (3), we also have
dP
s
b
d x =
−
This means that rate of change of P with respect to x depends on the quantity
s – b, which is the difference of selling price and the variable cost of each product Thus, in order to gain profit, this should be positive and to get large gains, we need to
produce large quantity of the product and at the same time try to reduce the variable
cost Example 7 Let a tank contains 1000 litres of brine which contains 250 g of salt per
litre |
1 | 306-309 | Remark From (3), we also have
dP
s
b
d x =
−
This means that rate of change of P with respect to x depends on the quantity
s – b, which is the difference of selling price and the variable cost of each product Thus, in order to gain profit, this should be positive and to get large gains, we need to
produce large quantity of the product and at the same time try to reduce the variable
cost Example 7 Let a tank contains 1000 litres of brine which contains 250 g of salt per
litre Brine containing 200 g of salt per litre flows into the tank at the rate of 25 litres per
minute and the mixture flows out at the same rate |
1 | 307-310 | Thus, in order to gain profit, this should be positive and to get large gains, we need to
produce large quantity of the product and at the same time try to reduce the variable
cost Example 7 Let a tank contains 1000 litres of brine which contains 250 g of salt per
litre Brine containing 200 g of salt per litre flows into the tank at the rate of 25 litres per
minute and the mixture flows out at the same rate Assume that the mixture is kept
uniform all the time by stirring |
1 | 308-311 | Example 7 Let a tank contains 1000 litres of brine which contains 250 g of salt per
litre Brine containing 200 g of salt per litre flows into the tank at the rate of 25 litres per
minute and the mixture flows out at the same rate Assume that the mixture is kept
uniform all the time by stirring What would be the amount of salt in the tank at
any time t |
1 | 309-312 | Brine containing 200 g of salt per litre flows into the tank at the rate of 25 litres per
minute and the mixture flows out at the same rate Assume that the mixture is kept
uniform all the time by stirring What would be the amount of salt in the tank at
any time t Solution Step 1 The situation is easily identifiable |
1 | 310-313 | Assume that the mixture is kept
uniform all the time by stirring What would be the amount of salt in the tank at
any time t Solution Step 1 The situation is easily identifiable Step 2 Let y = y (t) denote the amount of salt (in kg) in the tank at time t (in minutes)
after the inflow, outflow starts |
1 | 311-314 | What would be the amount of salt in the tank at
any time t Solution Step 1 The situation is easily identifiable Step 2 Let y = y (t) denote the amount of salt (in kg) in the tank at time t (in minutes)
after the inflow, outflow starts Further assume that y is a differentiable function |
1 | 312-315 | Solution Step 1 The situation is easily identifiable Step 2 Let y = y (t) denote the amount of salt (in kg) in the tank at time t (in minutes)
after the inflow, outflow starts Further assume that y is a differentiable function When t = 0, i |
1 | 313-316 | Step 2 Let y = y (t) denote the amount of salt (in kg) in the tank at time t (in minutes)
after the inflow, outflow starts Further assume that y is a differentiable function When t = 0, i e |
1 | 314-317 | Further assume that y is a differentiable function When t = 0, i e , before the inflow–outflow of the brine starts,
y = 250 g × 1000 = 250 kg
Note that the change in y occurs due to the inflow, outflow of the mixture |
1 | 315-318 | When t = 0, i e , before the inflow–outflow of the brine starts,
y = 250 g × 1000 = 250 kg
Note that the change in y occurs due to the inflow, outflow of the mixture Now the inflow of brine brings salt into the tank at the rate of 5 kg per minute
(as 25 × 200 g = 5 kg) and the outflow of brine takes salt out of the tank at the rate of
25 1000
40
y
y
=
kg per minute (as at time t, the salt in the tank is 1000
y
kg) |
1 | 316-319 | e , before the inflow–outflow of the brine starts,
y = 250 g × 1000 = 250 kg
Note that the change in y occurs due to the inflow, outflow of the mixture Now the inflow of brine brings salt into the tank at the rate of 5 kg per minute
(as 25 × 200 g = 5 kg) and the outflow of brine takes salt out of the tank at the rate of
25 1000
40
y
y
=
kg per minute (as at time t, the salt in the tank is 1000
y
kg) Thus, the rate of change of salt with respect to t is given by
dy
dt = 5
−y40
(Why |
1 | 317-320 | , before the inflow–outflow of the brine starts,
y = 250 g × 1000 = 250 kg
Note that the change in y occurs due to the inflow, outflow of the mixture Now the inflow of brine brings salt into the tank at the rate of 5 kg per minute
(as 25 × 200 g = 5 kg) and the outflow of brine takes salt out of the tank at the rate of
25 1000
40
y
y
=
kg per minute (as at time t, the salt in the tank is 1000
y
kg) Thus, the rate of change of salt with respect to t is given by
dy
dt = 5
−y40
(Why )
or
1
40
dy
y
dt +
= 5 |
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