The inequalities are 3x + 2y ≤ 54, 4x + 5y ≤ 100, x ≥ 0 and y ≥ 0

Now graph the all of four constraints.

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•  Draw the coordinate plane.
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Now first inequality 3x + 2y ≤ 54.

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• Graph the line y = (- 3x/2) + 27
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•  Since the inequality symbol is ≤ the boundary is included the solution set.
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• Graph the boundary of the inequality 3x + 2y ≤ 54 with solid line.
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• To determine which half plane to be shaded use a test point in either half- plane.
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• A simple choice is (0, 0). Substitute x  = 0 and y  = 0 in original inequality
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3x + 2y ≤ 54

0 ≤ 54

The statement is true.

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• Since the statement is true , shade the region contain point (0,0).
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Similarly graph the other inequalities.

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• Second inequality 4x + 5y ≤ 100
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Test point (0, 0)

0  ≤ 100

Since the statement is true , shade the region contain point (0, 0).

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• Third inequality x ≥ 0
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Test point (1, 1)

1 ≥ 0

Since the statement is true , shade the region contain point (1, 1).

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• Forth inequality y ≥ 0
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Test point (1, 1)

1 ≥ 0

Since the statement is true , shade the region contain point (1, 1).

Graph

The feasible area looks like in the graph

From the graph the corner points are (0, 0) ,(0, 20),(10, 12),(18, 0)

The function z = 15x + 10y

The maximum value is 350 at (10,12).