Question

Maximize \[ Z = 20x + 3y, \] subject to the constraints: \[ 3x + 2y \leq 10, \quad x \geq 0, \quad y \geq 0. \]

Hint:

In linear programming, maximize (or minimize) the objective function by evaluating it at the vertices (corner points) of the feasible region.

Answer 1

We need to maximize \[ Z = 20x + 3y \] subject to the constraints. Step 1: Identify the feasible region defined by: \[ 3x + 2y \leq 10, \quad x \geq 0, \quad y \geq 0. \] Step 2: Find corner points of the feasible region: - When \( x = 0 \), \[ 3(0) + 2y \leq 10 \implies y \leq 5. \] Corner point: \( (0, 0) \) and \( (0, 5) \). - When \( y = 0 \), \[ 3x + 0 \leq 10 \implies x \leq \frac{10}{3}. \] Corner point: \( \left(\frac{10}{3}, 0\right) \). - Intersection point of \[ 3x + 2y = 10. \] Step 3: Evaluate \(Z\) at corner points: \[ Z(0, 0) = 20 \times 0 + 3 \times 0 = 0, \] \[ Z(0, 5) = 20 \times 0 + 3 \times 5 = 15, \] \[ Z\left(\frac{10}{3}, 0\right) = 20 \times \frac{10}{3} + 3 \times 0 = \frac{200}{3} \approx 66.67. \] Maximum value of \(Z = \frac{200}{3}\) at \(\left(\frac{10}{3}, 0\right)\).