Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.
The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points are as follows.
|
Corner point |
Z = 3x + 4y |
|
|
O(0, 0) |
0 |
|
|
A(4, 0) |
12 |
|
|
B(0, 4) |
16 |
→ Maximum |
Therefore, the maximum value of Z is 16 at the point B (0, 4).
Popular Questions of Class 12 Mathematics
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(i) Symmetric but neither reflexive nor transitive.
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(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive. - Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
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(ii) Relation R in the set N of natural numbers defined as
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(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
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(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
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- Q:- Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. - Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
- Q:-
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- Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
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1 Comment(s) on this Question
Thanku sir
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