Question

optimize the objective function below based on the feasible region shown in the graph above. give both a minimum and a maximum if the region is bound.

objective function: \\(z = 5x + 3y\\)
constraints:
\\(3x + 6y \le 48\\)
\\(5x + 3y \le 45\\)
\\(x \ge 0\\)
\\(y \ge 0\\)

the maximum value is at the point(s)
the minimum value is at the points(s)

Explanation:

Step1: Find the boundary intersection points

Solve the system of boundary equations:

$$
LATEXBLOCK0
$$

Multiplying the second equation by 2 gives:

$$10x + 6y = 90$$

Subtracting the first equation from this yields:

$$7x = 42 \implies x = 6$$

Substituting \(x = 6\) into the second equation:

$$5(6) + 3y = 45 \implies 3y = 15 \implies y = 5$$

The intersection point is \((6, 5)\).

Step2: Identify all corner points

Find the intercepts of the boundary lines with the axes:
For \(3x + 6y = 48\):
If \(x = 0\), then \(y = 8\).
If \(y = 0\), then \(x = 16\).

For \(5x + 3y = 45\):
If \(x = 0\), then \(y = 15\).
If \(y = 0\), then \(x = 9\).

Considering the constraints \(x \ge 0\), \(y \ge 0\), and the inequalities, the vertices of the bounded feasible region are:

$$(0, 0), (0, 8), (6, 5), (9, 0)$$

Step3: Evaluate the objective function

Calculate \(Z = 5x + 3y\) at each vertex:
At \((0, 0)\):

$$Z = 5(0) + 3(0) = 0$$

At \((0, 8)\):

$$Z = 5(0) + 3(8) = 24$$

At \((6, 5)\):

$$Z = 5(6) + 3(5) = 30 + 15 = 45$$

At \((9, 0)\):

$$Z = 5(9) + 3(0) = 45$$

Step4: Determine maximum and minimum

The maximum value is 45, occurring at \((9, 0)\) and \((6, 5)\). The minimum value is 0, occurring at \((0, 0)\).

Answer:

The maximum value is 45 at the point(s) (9, 0) and (6, 5)
The minimum value is 0 at the point(s) (0, 0)