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 point(s)

Explanation:

Step1: Identify the boundary lines

The boundary lines of the constraints are:

$$3x + 6y = 48 \implies x + 2y = 16$$

$$5x + 3y = 45$$

$$x = 0, \quad y = 0$$

Step2: Find the corner points

Find the intersection points of the boundary lines in the first quadrant:

1. Intersection of \(x = 0\) and \(y = 0\): \((0, 0)\)
2. Intersection of \(5x + 3y = 45\) and \(y = 0\): \((9, 0)\)
3. Intersection of \(x + 2y = 16\) and \(x = 0\): \((0, 8)\)
4. Intersection of \(x + 2y = 16\) and \(5x + 3y = 45\):

Multiply the first equation by 5:

$$5x + 10y = 80$$

Subtract the second equation:

$$7y = 35 \implies y = 5$$

Substitute \(y = 5\) back:

$$x + 2(5) = 16 \implies x = 6$$

So, the intersection point is \((6, 5)\).

Step3: Evaluate the objective function

Evaluate \(z = 5x + 3y\) at each corner point:

• At \((0, 0)\): \(z = 5(0) + 3(0) = 0\)
• At \((9, 0)\): \(z = 5(9) + 3(0) = 45\)
• At \((0, 8)\): \(z = 5(0) + 3(8) = 24\)
• At \((6, 5)\): \(z = 5(6) + 3(5) = 45\)

Step4: Determine maximum and minimum

The maximum value is 45, which occurs at \((9, 0)\) and \((6, 5)\) (and along the line segment connecting them). Since the input box asks for the point(s), we can specify \((9, 0)\) or both. The minimum value is 0 at \((0, 0)\).

Answer:

The maximum value is 45 at the point(s) (9, 0).
The minimum value is 0 at the point(s) (0, 0).