Maximise Z = 250x + 75y subject to the constraints:
5x + y ≤ 100
x + y ≤ 60
x ≥ 0, y ≥ 0
Answers
Answer:
6250
Step-by-step explanation:
x<= 10
y <= 50
maximize will be hen the sum yield more
hence x=10, y = 50
Answer:
The most cost of Z is 5000 when x = 20 and y = 0
Step-by-step explanation:
From the above question,
They have given :
Maximise Z = 250x + 75y subject to the constraints:
5x + y ≤ 100
x + y ≤ 60
x ≥ 0, y ≥ 0
To maximize Z = 250x + 75y issue to the given constraints:
We want to take a look at the feasibility of the given constraints.
x, y are each non-negative, so the constraint x ≥ 0, y ≥ zero is satisfied.
5x + y ≤ a hundred can be rewritten as y ≤ one hundred - 5x.
x + y ≤ 60 can be rewritten as y ≤ 60 - x.
Since y has to be much less than or equal to each of these expressions, we can mix them to get y ≤ min(100 - 5x, 60 - x).
Therefore, the possible vicinity for the given constraints is the set of all non-negative values of x and y that lie beneath the line y = min(100 - 5x, 60 - x).
To maximize Z, we want to locate the factor on the boundary of the viable area that offers the most price of Z.
The boundary of the possible area is made up of the line segments that structure the edges of the viable region.
To locate the most price of Z on the boundary, we want to consider Z at every of the endpoints of these line segments.
The endpoints of these line segments are the factors the place the line y = min(100 - 5x, 60 - x) intersects the coordinate axes.
We can discover these endpoints via putting y = zero and fixing for x, and placing x = zero and fixing for y.
Solving for y = 0 offers us the factors (20, 0) and (0, 60).
Solving for x = 0 offers us the factor (0, 0).
Evaluating Z at these factors offers us Z(20, 0) = 5000, Z(0, 60) = 4500, and Z(0, 0) = 0.
Therefore, the most price of Z is 5000, which happens at the factor (20, 0).
Hence, the most cost of Z is 5000 when x = 20 and y = 0
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