Question

graph each system of inequalities representing the constraints for the objective function below. then determine what the maximum value is and the coordinate values of x and y for which the maximum occurs. you can use desmos graphing calculator to help you. watch this video if you need help with this problem. objective function z = 6x + 10y constraints \

$$\begin{cases} x + y \\leq 12 \\\\ x + 2y \\leq 20 \\\\ \\begin{cases} x \\geq 0 \\\\ y \\geq 0 \\end{cases}$$

\text{quadrant i and its boundary} \end{cases} maximum: ---- coordinate: ----

Explanation:

Step1: Find Feasible Region Vertices

First, identify the intersection points of the constraint lines (including axes, since \(x \geq 0, y \geq 0\)):

• Intersection of \(x = 0\) and \(x + y = 12\): \(y = 12\), so \((0, 12)\).
• Intersection of \(x + y = 12\) and \(x + 2y = 20\): Subtract equations: \(y = 8\), then \(x = 4\), so \((4, 8)\).
• Intersection of \(x + 2y = 20\) and \(y = 0\): \(x = 20\), but check \(x + y \leq 12\): \(20 + 0 > 12\), so invalid. Instead, intersection of \(x + 2y = 20\) and \(x = 0\): \(y = 10\), but \(0 + 10 = 10 \leq 12\), wait—no, \(x + y \leq 12\) and \(x + 2y \leq 20\). Wait, correct vertices in Quadrant I:
• \((0, 0)\) (origin), but check constraints: \(0 + 0 \leq 12\), \(0 + 0 \leq 20\), valid.
• \((0, 10)\)? No, \(x + y = 0 + 10 = 10 \leq 12\), but \(x + 2y = 20\) at \(x=0\) is \(y=10\), so \((0, 10)\)? Wait, no: solve \(x=0\) in \(x + 2y \leq 20\): \(y \leq 10\), and \(x + y \leq 12\): \(y \leq 12\). So stricter is \(y \leq 10\). But also, intersection of \(x + y = 12\) and \(x=0\) is \((0,12)\), but \(x + 2y = 24 > 20\), so \((0,12)\) is invalid (violates \(x + 2y \leq 20\)). So correct vertices:

1. \((0, 0)\): Check \(z = 0\).
2. Intersection of \(x=0\) and \(x + 2y = 20\): \((0, 10)\) (since \(0 + 2(10) = 20\), and \(0 + 10 = 10 \leq 12\), valid).
3. Intersection of \(x + y = 12\) and \(x + 2y = 20\): As before, \(y = 8\), \(x = 4\) → \((4, 8)\).
4. Intersection of \(x + y = 12\) and \(y = 0\): \((12, 0)\) (check \(12 + 0 = 12 \leq 20\), valid).
5. \((0, 0)\) is valid, but let's confirm all vertices:

• \((0, 0)\): \(z = 0\).
• \((0, 10)\): \(z = 6(0) + 10(10) = 100\).
• \((4, 8)\): \(z = 6(4) + 10(8) = 24 + 80 = 104\).
• \((12, 0)\): \(z = 6(12) + 10(0) = 72\).

Step2: Evaluate Objective Function

Evaluate \(z = 6x + 10y\) at each vertex:

• \((0, 0)\): \(z = 0\).
• \((0, 10)\): \(z = 100\).
• \((4, 8)\): \(z = 6(4) + 10(8) = 24 + 80 = 104\).
• \((12, 0)\): \(z = 72\).

Answer:

Maximum value: \(104\)
Coordinate: \((4, 8)\)