The objective function Z = 2x + 3y cannot be maximized, while the maximum value of the objective function Z = 2x + 5y is 19
<h3>Part A (Question 1): Maximize Z</h3>The objective function is given as:
Z = 2x + 3y
The constraints are given as:
3x+2y≤15
4x+y≤0
x≥0; y≥0
See attachment for the graph of the constraints
From the attached graph, we have the optimal solution to be:
(x,y) = (-3,12)
Because x≥0 & y≥0, and x is negative in (-3,12), then
The function cannot be maximized.
<h3>Part B (Question 7): Maximize Z</h3>The objective function is given as:
Z = 2x + 5y
The constraints are given as:
x + y ≤ 5
y - x ≤ 1
x≥0; y≥0
See attachment for the graph of the constraints
From the attached graph, we have the optimal solution to be:
(x,y) = (2,3)
Substitute 2 and 3 for x and y in the objective function.
Z = 2 * 2 + 5 * 3
Evaluate
Z = 19
Hence, the maximum value of the objective function is 19
The questions 21, 22, 23 and 25 are not included in the attachment
Read more about maximized functions at:
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bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of the provided line then
well, we know the x-intercept is at x = 3, recall when a graph intercepts the x-axis y = 0, so this point is (3 , 0). Then we're really looking for the equation of a line whose slope is 1/5 and runs through (3 , 0).
