Find the values of \( x \geq 0 \) and \( y \geq 0 \) that maximize \( z=11 x+14 y \), subject to each of the following sets of constraints. \( \begin{array}{lll}\text { (a) } \begin{array}{rr}x+y \leq 16 & \text { (b) } 2 x+y \leq 10\end{array} & \text { (c) } x+2 y \geq 8 \\ x+5 y \leq 20 & \begin{array}{r}2 x+y \geq 10 \\ x+5 y \leq 15\end{array} \\ x-y \leq 6\end{array} \) (a) Select the correct choice and, if necessary, fill in the answer boxes to complete your choice. A. The maximum value is \( \square \) and occurs at the point (Simplify your answers.) B. There is no maximum value.

To solve the problem of maximizing \( z = 11x + 14y \) under the given constraints, we will analyze each set of constraints step by step. ### Part (a) **Constraints:** 1. \( x + y \leq 16 \) 2. \( x + 5y \leq 20 \) 3. \( x \geq 0 \) 4. \( y \geq 0 \) **Step 1: Graph the constraints** 1. From \( x + y \leq 16 \): - The line is \( y = 16 - x \). - Intercepts: \( (16, 0) \) and \( (0, 16) \). 2. From \( x + 5y \leq 20 \): - The line is \( y = \frac{20 - x}{5} \). - Intercepts: \( (20, 0) \) and \( (0, 4) \). **Step 2: Find the feasible region** The feasible region is bounded by the lines and the axes. We will find the intersection points of the lines: 1. Set \( x + y = 16 \) and \( x + 5y = 20 \): \[ x + y = 16 \quad (1) \\ x + 5y = 20 \quad (2) \] Substituting \( x = 16 - y \) into (2): \[ 16 - y + 5y = 20 \\ 16 + 4y = 20 \\ 4y = 4 \\ y = 1 \\ x = 16 - 1 = 15 \] Intersection point: \( (15, 1) \). **Step 3: Evaluate the objective function at the vertices of the feasible region** Vertices of the feasible region: 1. \( (0, 0) \) 2. \( (0, 4) \) 3. \( (16, 0) \) 4. \( (15, 1) \) Calculating \( z \) at each vertex: 1. \( z(0, 0) = 11(0) + 14(0) = 0 \) 2. \( z(0, 4) = 11(0) + 14(4) = 56 \) 3. \( z(16, 0) = 11(16) + 14(0) = 176 \) 4. \( z(15, 1) = 11(15) + 14(1) = 165 + 14 = 179 \) **Step 4: Determine the maximum value** The maximum value occurs at \( (15, 1) \) with \( z = 179 \). ### Conclusion for Part (a) A. The maximum value is \( 179 \) and occurs at the point \( (15, 1) \). Now, let's proceed to part (b) and (c) in a similar manner. ### Part (b) **Constraints:** 1. \( 2x + y \leq 10 \) 2. \( 2x + y \geq 10 \) 3. \( x + 5y \leq 15 \) 4. \( x - y \leq 6 \) 5. \( x \geq 0 \) 6. \( y \geq 0 \) **Step 1: Graph the constraints** 1. From \( 2x + y = 10 \): - Intercepts: \( (5, 0) \) and \( (0, 10) \). 2. From \( x + 5y = 15 \): - Intercepts: \( (15, 0) \) and \( (0, 3) \). 3. From \( x - y = 6 \): - Intercepts: \( (6, 0) \) and \( (0, -6) \) (not relevant since \( y \geq 0 \)). **Step 2: Find the feasible region** The feasible region is bounded by the lines and the axes. We will find the intersection points of the lines: 1. Set \( 2x + y = 10 \) and \( x + 5y = 15 \): \[ 2x + y = 10 \quad (1) \\ x + 5y = 15 \quad (2) \] Substituting \( y = 10 - 2x \) into (2): \[ x + 5(10 - 2x) = 15 \\ x + 50 - 10x = 15 \\ -9x = -35 \\ x = \frac{35}{9} \\ y = 10 - 2\left(\frac{35}{9}\right) = \frac{10}{9} \] Intersection point: \( \left(\frac{35}{9}, \frac{10}{9}\right) \). **Step 3: Evaluate the objective function at the vertices of the feasible region** Vertices of the feasible region: 1. \( (0, 0) \) 2. \( (5, 0) \) 3. \( (0, 10) \) 4. \( (6, 0) \) 5. \( \left(\frac{35}{9}, \frac{10}{9}\right) \) Calculating \( z \) at each vertex: 1. \( z(0, 0) = 0 \) 2. \( z(5, 0) = 55 \) 3. \( z(0, 10) = 140 \) 4. \( z(6, 0) = 66 \) 5. \( z\left(\frac{35}{9}, \frac{10}{9}\right) = 11\left(\frac{35}{9}\right) + 14\left(\frac{10}{9}\right) = \frac{385}{9} + \frac{140}{9} = \frac{525}{9} \approx 58.33 \) **Step 4: Determine the maximum value** The maximum value occurs at \( (0, 10) \) with \( z = 140 \). ### Conclusion for Part (b) A. The maximum value is \( 140 \) and occurs at the point \( (0, 10) \). ### Part (c) **Constraints:** 1. \( x + 2y \geq 8 \) 2. \( 2x + y \geq 10 \) 3. \( x + 5y \leq 15 \) 4. \( x - y \leq 6 \) 5. \( x \geq 0 \) 6