Maximize the objective function p = 5x + 3y for the given constraints. X ≥ 0 y ≥ 0 3x + 6y ≤ 48 2x + 8y ≤ 56.

Answers

Answer 1

The maximized value of the objective function is 38

The objective function is given as:

Max P = 5x + 3y

The constraints are given as:

3x + 6y ≤ 48 2x + 8y ≤ 56x ≥ 0 y ≥ 0

To do this, we make use of the graphical method.

See attachment for the graph of the constraints.

From the graph, we have the following feasible point

[tex](x,y) = (4,6)[/tex]

Substitute 4 for x and 6 for y in the objective function.

[tex]Max\ P = 5x + 3y[/tex]

So, we have:

[tex]P = 5 \times 4 + 3 \times 6[/tex]

[tex]P = 20 + 18[/tex]

[tex]P = 38[/tex]

Hence, the maximized value of the objective function is 38

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Maximize The Objective Function P = 5x + 3y For The Given Constraints. X 0 Y 0 3x + 6y 48 2x + 8y 56.
Answer 2

Answer:

Explanation:

If an objective function has a maximum value, it will occur at one of the vertices of the feasible region (i.e., the solution region for the system of inequalities). The feasible region's vertices occur at (0, 0), (0, 7), (4, 6), and (16, 0).

Use the x- and y-value from each vertex to evaluate the objective function P = 5x + 3y.

(x, y) 5x + 3y P

(0, 0) 5(0) + 3(0) 0

(0, 7) 5(0) + 3(7) 21

(4, 6) 5(4) + 3(6) 38

(16, 0) 5(16) + 3(0) 80

Therefore, the maximum of the objective function is 80.


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