Download Exercise 13B

13
NON-FOUNDATION
Linear Inequalities in Two
Unknowns
Name :
13B
Date :
Mark :
13.2 Linear Programming
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In each of the following figures, (1 – 2)
(a) write down the system of inequalities that determines the shaded region,
(b) find the maximum and the minimum values of P = x - y if (x, y) is any point in the shaded
region.
1.
y
4
x-y=0
2
8x + 3y = 20
-3x + 2y = 5
-3
-2
x
-1
0
1
-2
2
3
4
6x + 5y = 4
-4
Solution
(a) From the shaded region, take (1, 0) as the test point.
For the line 8x + 3y = 20, evaluate the value of 8x + 3y for the point (1, 0).
Q
8( 1
\
8x + 3 y ( ≥
) + 3( 0
/
) = ( 8 )
( ≥ / £ ) 20
£ ) 20 is one of the inequalities.
For the line 6x + 5y = 4, evaluate the value of 6x + 5y for the point (1, 0).
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Q
6( 1
\
6x + 5 y ( ≥
) + 5( 0
/
) = ( 6 )
( ≥ / £ )4
£ ) 4 is one of the inequalities.
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13
Linear Inequalities in Two Unknowns
For the line -3x + 2y = 5, evaluate the value of -3x + 2y for the point (1, 0).
Q
-3( 1
\
-3x + 2 y ( ≥
\
Ï8 x + 3 y ( ≥ / £ ) 20
Ô
The system of inequalities is Ì6 x + 5 y ( ≥ / £ ) 4 .
Ô-3x + 2 y ( ≥ / £ ) 5
Ó
) + 2(
0
/
) = ( -3 )
( ≥ / £ )5
£ ) 5 is one of the inequalities.
(b) Draw the line x - y = 0 on the given graph.
Translate the line x - y = 0 in the ( positive / negative ) direction of the x-axis to
obtain increasing values of P.
From the graph, P attains its maximum at ( 4 , -4 ).
\ Maximum value of P = ( 4 ) - ( -4 ) = ( 8 )
Translate the line x - y = 0 in the ( positive / negative ) direction of the x-axis to obtain
decreasing values of P.
From the graph, P attains its minimum at ( 1 , 4 ).
\ Minimum value of P = ( 1 ) - ( 4 ) = ( -3 )
2.
y
2x - 5y = -6
2
1
4x + 7y = 9
x
-3
-2
x -y = 0
-1
0
1
-1
2
3
4
x + 7y = -3
Solution
(a) From the shaded region, take (0, 0) as the test point.
Evaluate the value of 4x + 7y for the point (0, 0).
Q
4(0) + 7(0) = 0
£ 9
\
4x + 7y £ 9 is one of the inequalities.
(Solution continues on the next page.)
55
Number and Algebra
Evaluate the value of 2x - 5y for the point (0, 0).
Q
2(0) - 5(0) = 0
≥ -6
\
2x - 5y ≥ -6 is one of the inequalities.
Evaluate the value of x + 7y for the point (0, 0).
Q
(0) + 7(0) = 0
≥ -3
\
x + 7y ≥ -3 is one of the inequalities.
\
Ï4x + 7y £ 9
Ô
The system of inequalities is Ì2x - 5y ≥ -6 .
Ô
Óx + 7y ≥ -3
(b) Draw the line x - y = 0 on the given graph.
Translate the line x - y = 0 in the positive direction of the x-axis to obtain increasing
values of P.
From the graph, P attains its maximum at ( 4 , -1 ).
\
Maximum value of P = 4 - (-1) = 5
Translate the line x - y = 0 in the negative direction of the x-axis to obtain decreasing
values of P.
From the graph, P attains its minimum at ( -3 , 0 ).
\
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Minimum value of P = (-3) - 0 = -3
13
Linear Inequalities in Two Unknowns
In each of the following, find the maximum and the minimum values of P subject to the given
constraints. (3 – 4)
3.
P = 2x - 4 y
Ïx £ 1
Ô
Ì x - 4 y ≥ -7
Ô3x + 2 y ≥ -7
Ó
Solution
Shade the region that satisfies the following constraints:
Ïx £ 1
Ô
Ì x - 4 y ≥ -7
Ô3x + 2 y ≥ -7
Ó
x - 4 y = -7
3x + 2 y = -7
x
-3
-1
1
x
-3
-1
1
y
1
1.5
2
y
1
-2
-5
x − 4y = −7
y
2
1
x
−3
−2
−1
0
−1
2x − 4y = 0
3x + 2y = −7
1
2
x=1
−2
−3
−4
−5
Draw the line 2 x - 4 y = 0 on the above graph.
(Solution continues on the next page.)
57
Number and Algebra
Translate the line 2 x - 4 y = 0 in the ( positive / negative ) direction of the x-axis to obtain
increasing values of P.
From the graph, P attains its maximum at ( 1
Maximum value of P = 2(
\
1
, -5 ).
) - 4( -5
) = ( 22
)
Translate the line 2 x - 4 y = 0 in the ( positive / negative ) direction of the x-axis to obtain
decreasing values of P.
From the graph, P attains its minimum at ( -3 , 1
Minimum value of P = 2( -3 ) - 4(
\
4.
) = ( -10 )
1
P = x - 3y
Ï3x + 5 y ≥ 5
Ô
Ì7 x - 5 y ≥ -8
Ô y ≥ 12 x - 62
Ó
Solution
Shade the region that satisfies the constraints.
3x + 5y = 5
7x - 5y = -8
x
-5
0
5
x
-4
1
6
y
4
1
-2
y
-4
3
10
y = 12x - 62
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x
5
5.5
6
y
-2
4
10
).
13
Linear Inequalities in Two Unknowns
y
12
10
8
7x − 5y = −8
6
y = 12x − 62
4
3x + 5y = 5
2
x − 3y = 0
x
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
7
−2
−4
Draw the line x - 3y = 0 on the above graph.
Translate the line x - 3y = 0 in the positive direction of the x-axis to obtain increasing values of P.
From the graph, P attains its maximum at ( 5 , -2 ).
\ Maximum value of P = 5 - 3(-2) = 11
Translate the line x - 3y = 0 in the negative direction of the x-axis to obtain decreasing values of P.
From the graph, P attains its minimum at ( 6 , 10 ).
\ Minimum value of P = 6 - 3(10) = -24
59
Number and Algebra
5.
(a) Shade the region that satisfies all the following constraints:
Ï8 x + 6 y £ 31
Ô
Ì6 x - 8 y ≥ -33
Ô2 x + 4 y ≥ 9
Ó
(b) Find the maximum and the minimum values of P = 3x + 2 y subject to the constraints in
(a) if x and y are
real numbers,
(i)
(ii) integers.
Solution
(a) 8 x + 6 y = 31
6 x - 8 y = -33
x
0.5
2
3.5
x
-1.5
0.5
2.5
y
4.5
2.5
0.5
y
3
4.5
6
x
-1.5
0.5
2.5
y
3
2
1
2x + 4 y = 9
y
6
6x - 8y = -33
8x + 6y = 31
5
4
3
2
1
3x + 2y = 0
2x + 4y = 9
x
-2
60
-1
0
1
2
3
4
13
Linear Inequalities in Two Unknowns
Draw the line 3x + 2 y = 0 on the graph in (a).
(b) (i)
From the graph, P attains its maximum at ( 3.5 , 0.5 ).
\ Maximum value of P = 3( 3.5 ) + 2( 0.5 ) = ( 11.5 )
From the graph, P attains its minimum at ( -1.5 , 3
\ Minimum value of P = 3( -1.5 ) + 2(
).
) = ( 1.5 )
3
(ii) Mark black dots on the same graph to show all the feasible solutions with integral xand y- coordinates.
From the graph, P attains its maximum at (
\ Maximum value of P = 3(
3
, 1
3
) + 2(
1
) = ( 11 )
From the graph, P attains its minimum at ( -1 , 3
\ Minimum value of P = 3( -1 ) + 2(
6.
3
).
).
) = (
3
)
(a) Shade the region that satisfies all the following constraints:
Ï6 x + 8 y + 7 £ 0
Ô
Ì10 x + 4 y + 21 ≥ 0
Ô2 x + 12 y + 21 ≥ 0
Ó
(b) Find the maximum and the minimum values of P = 3x - y + 2 subject to the
constraints in (a) if x and y are
real numbers,
(i)
(ii) integers.
Solution
(a) 6x + 8y + 7 = 0
x
y
10x + 4y + 21 = 0
-2.5 -0.5 1.5
x
-0.5 -2
y
1
-2.5 -1.5 -0.5
1
-1.5 -4
2x + 12y + 21 = 0
x
-1.5 1.5
4.5
y
-1.5 -2
-2.5
(Solution continues on the next page.)
61
Number and Algebra
y
10x + 4y + 21 = 0
3x − y + 2 = 0
2
1
x
−3
−2
−1
0
1
2
3
4
−1
−2
2x + 12y + 21 = 0
−3
−4
(b) (i)
6x + 8y + 7 = 0
Draw the line 3x - y + 2 = 0 on the graph in (a).
From the graph, P attains its maximum at ( 1.5 , -2 ).
\ Maximum value of P = 3(1.5) - (-2) + 2 = 8.5
From the graph, P attains its minimum at ( -2.5 , 1 ).
\ Minimum value of P = 3(-2.5) - 1 + 2 = -6.5
(ii) Mark black dots on the same graph to show all the feasible solutions with integral xand y- coordinates.
From the graph, P attains its maximum at ( 0 , -1 ).
\ Maximum value of P = 3(0) - (-1) + 2 = 3
From the graph, P attains its minimum at ( -2 , 0 ).
\ Minimum value of P = 3(-2) - 0 + 2 = -4
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