Question

using the geometric approach, we want to maximize the objective function z = 6x + 13y subject to the following constraints:
\\(\

$$\begin{cases}6x + 2y\\geq36 \\\\ 2x + 4y\\geq32 \\\\ y\\leq20 \\\\ x\\leq10 \\\\ x\\geq0 \\\\ y\\geq0\\end{cases}$$

\\)
the graph below shows the boundary lines for the above constraint inequalities but is not shaded.
select the one answer that contains the correct coordinates of the corner points of what must be shaded as the solution region, and identify the associated maximum value given the above objective function, z.
solution region corner points: (0, 0), (0, 8), (2, 6), (6, 0)
z = 104 at maximum
solution region corner points: (0, 0), (0, 8), (10, 3), (10, 0)
z = 104 at maximum
solution region corner points: (10, 3), (4, 6), (0, 18), (0, 20), (10, 20)
z = 320 at maximum
solution region corner points: (0, 0), (0, 8), (20, 0)
z = 120 at maximum
solution region corner points: (0, 18), (0, 8), (2, 6)
z = 104 at maximum
correct answer is not listed
solution region corner points: (0, 0), (10, 0), (0, 18), (0, 20), (10, 20)
z = 320 at maximum

Explanation:

Step1: Identify constraint lines

Constraints: \(6x+2y \geq 36\), \(2x+4y \geq 32\), \(y \leq 20\), \(x \leq 10\), \(x,y \geq 0\).

Step2: Find intercepts/intersections

• \(6x+2y=36\): intercepts \((6,0)\), \((0,18)\); intersects \(2x+4y=32\) at \((4,6)\).
• \(2x+4y=32\): intercepts \((16,0)\), \((0,8)\); intersects \(x=10\) at \((10,3)\).
• Bounds: \(x=10\) & \(y=20\) intersect at \((10,20)\); \(y=20\) & \(x=0\) at \((0,20)\).

Step3: Determine feasible corner points

Feasible region corners: \((0,18)\), \((4,6)\), \((10,3)\), \((10,20)\), \((0,20)\).

Step4: Calculate Z at corners

\(Z=6x+13y\):

• \((0,18)=234\), \((4,6)=102\), \((10,3)=99\), \((10,20)=320\), \((0,20)=260\). Max Z=320.

Answer:

Solution region corner points: (10, 3) , (4, 6) , (0, 18) , (0, 20) , (10, 20), Z = 320 at maximum