Using the following system of inequalities find the maximum value of f(x,y) = 3x + 5y x≥0x≥0 y≥0y≥0 3x+2y≤183x+2y≤18 6x+7y≤426x+
7y≤42
1 answer:
Given the objective function f(x, y) = 3x + 5y
subject to the constraints:
<span>x≥0
y≥0
3x+2y≤18
6x+7y≤42
The graphical representation of the contraints is attached.
From the graph, it can be seen that the corner ponts of the graph are (0, 0), (0, 6), (4.667, 2), (6, 0)
The values of the objective function at the various corner points are as follows:
For (0, 0): f(x, y) = 3(0) + 5(0) = 0 + 0 = 0
For (0, 6): f(x, y) = 3(0) + 5(6) = 0 + 30 = 30
For (4.667, 2): f(x, y) = 3(4.667) + 5(2) = 14 + 10 = 24
For (6, 0): f(x, y) = 3(6) + 5(0) = 18 + 0 = 18
From the above, it can be seen that the maximum value of the objective function is 30 and it occurs when x = 0 and y = 6.
Therefore, The maximum value of the given system is 30.</span>
subject to the constraints:
<span>x≥0
y≥0
3x+2y≤18
6x+7y≤42
The graphical representation of the contraints is attached.
From the graph, it can be seen that the corner ponts of the graph are (0, 0), (0, 6), (4.667, 2), (6, 0)
The values of the objective function at the various corner points are as follows:
For (0, 0): f(x, y) = 3(0) + 5(0) = 0 + 0 = 0
For (0, 6): f(x, y) = 3(0) + 5(6) = 0 + 30 = 30
For (4.667, 2): f(x, y) = 3(4.667) + 5(2) = 14 + 10 = 24
For (6, 0): f(x, y) = 3(6) + 5(0) = 18 + 0 = 18
From the above, it can be seen that the maximum value of the objective function is 30 and it occurs when x = 0 and y = 6.
Therefore, The maximum value of the given system is 30.</span>
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