Question

Minimize \[ Z = 2x + y, \] subject to \[ 5x + 10y \le 50, \quad x + y \ge 1, \quad y \le 4, \quad x \ge 0, \quad y \ge 0. \]

Hint:

Identify the feasible region, find vertices by solving constraint equations, and evaluate the objective function at vertices to find min or max values.

Answer 1

Step 1: Write down the constraints: \[ 5x + 10y \le 50 \implies x + 2y \le 10, \] \[ x + y \ge 1, \] \[ y \le 4, \] \[ x \ge 0, \quad y \ge 0. \] Step 2: Identify the feasible region defined by these inequalities. Step 3: Find the corner points (vertices) of the feasible region by solving the boundary equations: - Intersection of \(x + 2y = 10\) and \(x + y = 1\): \[ x + 2y = 10, \] \[ x + y = 1 \implies x = 1 - y. \] Substitute: \[ 1 - y + 2y = 10 \implies y = 9, \quad x = 1 - 9 = -8 \quad (\text{Not feasible since } x \ge 0). \] - Intersection of \(x + 2y = 10\) and \(y = 4\): \[ x + 2(4) = 10 \implies x = 10 - 8 = 2. \] Point: \((2,4)\) (feasible since \(x,y \ge 0\)). - Intersection of \(x + y = 1\) and \(y = 4\): \[ x + 4 = 1 \implies x = -3 \quad (\text{Not feasible}). \] - Intersection of \(x + y = 1\) and \(y = 0\): \[ x + 0 = 1 \implies x = 1, \] Point: \((1,0)\). - Intersection of \(x + 2y = 10\) and \(y = 0\): \[ x + 0 = 10 \implies x = 10, \] Point: \((10,0)\). - Check \(y \le 4\), \(x \ge 0\), \(y \ge 0\). Step 4: Evaluate \(Z = 2x + y\) at feasible corner points: \[ (1,0) \to Z = 2(1) + 0 = 2, \] \[ (2,4) \to Z = 2(2) + 4 = 8, \] \[ (10,0) \to Z = 2(10) + 0 = 20, \] \[ (0,1) \text{ (from } x + y \ge 1 \text{ with } x=0) \to Z = 0 + 1 = 1, \] \[ (0,0) \text{ (check feasibility) violates } x + y \ge 1. \] Step 5: The minimum value of \(Z\) in the feasible region is at \((0,1)\) with \[ \boxed{Z_{\min} = 1}. \]