1. Rephrase inequalities as equations for graphing: - \( x + 2y = 4 \) is rewritten as \( y = \frac{4 - x}{2} \). - \( 3x + y = 6 \) is rewritten as \( y = 6 - 3x \). - \( x + y = 4 \) is rewritten as \( y = 4 - x \). 2. Identify the feasible region: - The feasible region is the intersection of the half-planes defined by: - \( x + 2y \geq 4 \): Region above \( x + 2y = 4 \). - \( 3x + y \leq 6 \): Region below \( 3x + y = 6 \). - \( x + y \leq 4 \): Region below \( x + y = 4 \). - \( x \geq 0, y \geq 0 \): The first quadrant. Graph these constraints and shade the area satisfying all inequalities.3. Determine the corner points of the feasible region: Corner points are found by solving for the intersection of boundary lines: - Intersection of \( x + 2y = 4 \) and \( 3x + y = 6 \): Solving the system: \[ x + 2y = 4 \quad \text{and} \quad 3x + y = 6. \] From \( x + 2y = 4 \), \( y = \frac{4 - x}{2} \). Substituting into \( 3x + y = 6 \): \[ 3x + \frac{4 - x}{2} = 6. \] Simplifying: \[ \frac{6x + 4 - x}{2} = 6 \implies \frac{5x + 4}{2} = 6 \implies 5x + 4 = 12 \implies x = \frac{8}{5}. \] Substituting \( x = \frac{8}{5} \) into \( y = \frac{4 - x}{2} \): \[ y = \frac{4 - \frac{8}{5}}{2} = \frac{\frac{20}{5} - \frac{8}{5}}{2} = \frac{\frac{12}{5}}{2} = \frac{6}{5}. \] Corner point: \( \left(\frac{8}{5}, \frac{6}{5}\right) \). - Intersection of \( x + 2y = 4 \) and \( x + y = 4 \): Solving the system: \[ x + 2y = 4 \quad \text{and} \quad x + y = 4. \] From \( x + y = 4 \), \( y = 4 - x \). Substituting into \( x + 2y = 4 \): \[ x + 2(4 - x) = 4 \implies x + 8 - 2x = 4 \implies -x + 8 = 4 \implies x = 4. \] Substituting \( x = 4 \) into \( y = 4 - x \): \[ y = 4 - 4 = 0. \] Corner point: \( (4, 0) \). - Intersection of \( 3x + y = 6 \) and \( x + y = 4 \): Solving the system: \[ 3x + y = 6 \quad \text{and} \quad x + y = 4. \] Subtracting \( x + y = 4 \) from \( 3x + y = 6 \): \[ 3x + y - (x + y) = 6 - 4 \implies 2x = 2 \implies x = 1. \] Substituting \( x = 1 \) into \( x + y = 4 \): \[ y = 4 - 1 = 3. \] Corner point: \( (1, 3) \).4. Evaluate \( z = 5x + 4y \) at the corner points: - At \( \left(\frac{8}{5}, \frac{6}{5}\right) \): \[ z = 5\left(\frac{8}{5}\right) + 4\left(\frac{6}{5}\right) = 8 + \frac{24}{5} = \frac{40}{5} + \frac{24}{5} = \frac{64}{5}. \] - At \( (4, 0) \): \[ z = 5(4) + 4(0) = 20. \] - At \( (1, 3) \): \[ z = 5(1) + 4(3) = 5 + 12 = 17. \]5. Determine the optimal values: - The maximum value of \( z \) is \( 20 \) occurring at \( (4, 0) \). - The minimum value of \( z \) is \( 17 \) occurring at \( (1, 3) \). Final Answer: - Maximum value: \( z = 20 \) at \( (4, 0) \). - Minimum value: \( z = 17 \) at \( (1, 3) \).