Let's solve the given linear programming problem **step by step**: ### 1. **Write the constraints:** Given: \[ \begin{align*} \text{Maximize:} \quad & p = 3x + y \\ \text{Subject to:} \quad & 3x - 7y \leq 0 \\ & 7x - 3y \geq 0 \\ & x + y \leq 10 \\ & x \geq 0, \; y \geq 0 \end{align*} \] --- ### 2. **Rewrite constraints for graphing:** 1. \( 3x - 7y \leq 0 \implies 3x \leq 7y \implies x \leq \frac{7}{3}y \) 2. \( 7x - 3y \geq 0 \implies 7x \geq 3y \implies x \geq \frac{3}{7}y \) 3. \( x + y \leq 10 \) 4. \( x \geq 0 \) 5. \( y \geq 0 \) --- ### 3. **Find the corner points (vertices):** Let's solve the system by finding intersections: #### **A. Intersection of \( x = \frac{7}{3}y \) and \( x + y = 10 \):** \[ x = \frac{7}{3}y \\ \frac{7}{3}y + y = 10 \\ \frac{10}{3}y = 10 \implies y = 3,\ x = \frac{7}{3}(3) = 7 \] **Point 1:** \((7, 3)\) --- #### **B. Intersection of \( x = \frac{3}{7}y \) and \( x + y = 10 \):** \[ x = \frac{3}{7}y \\ \frac{3}{7}y + y = 10 \\ \frac{10}{7}y = 10 \implies y = 7,\ x = \frac{3}{7}(7) = 3 \] **Point 2:** \((3, 7)\) --- #### **C. Intersection of \( x = \frac{3}{7}y \) and \( x = \frac{7}{3}y \):** \[ \frac{3}{7}y = \frac{7}{3}y \implies 3^2 y = 7^2 y \implies 9y = 49y \implies y = 0 \\ x = 0 \] **Point 3:** \((0, 0)\) --- #### **D. Intersection with axes:** - For \( x + y = 10 \): \( (10,0) \) and \( (0,10) \) - For \( x = \frac{3}{7}y \): At \( y = 0 \Rightarrow x = 0 \) - For \( x = \frac{7}{3}y \): At \( y = 0 \Rightarrow x = 0 \) --- #### **Check which points are feasible:** Test the points in all constraints: - \((0, 0)\): Feasible. - \((7, 3)\): Check \( x \leq \frac{7}{3}y \implies 7 \leq 7 \) (OK), \( x \geq \frac{3}{7}y \implies 7 \geq 1.29 \) (OK), \( 7+3=10 \leq 10 \) (OK) - \((3, 7)\): Check \( x \leq \frac{7}{3}y \implies 3 \leq 16.33 \) (OK), \( x \geq \frac{3}{7}y \implies 3 \geq 3 \) (OK), \( 3+7=10 \leq 10 \) (OK) Check \((10,0)\): - \( x \leq \frac{7}{3}y \implies 10 \leq 0 \) (**not OK**) Check \((0,10)\): - \( x \geq \frac{3}{7}y \implies 0 \geq 4.285 \) (**not OK**) --- ### 4. **Evaluate the Objective Function at the Feasible Points** \[ p = 3x + y \] - At **(0,0)**: \( p = 0 \) - At **(7,3)**: \( p = 3(7) + 3 = 21 + 3 = 24 \) - At **(3,7)**: \( p = 3(3) + 7 = 9 + 7 = 16 \) --- ### 5. **Final Answer** \[ \boxed{p = 24 \text{ at } (x, y) = (7, 3)} \] --- ### **Graph Sketch (Description)** - Draw the lines \( x = \frac{7}{3}y \) and \( x = \frac{3}{7}y \). - Draw the line \( x + y = 10 \). - Shade the feasible region bounded by these lines in the first quadrant (\( x \geq 0, y \geq 0 \)). - The **optimal (maximum) point is at (7, 3)**. --- **If you want an actual graph, let me know!**