Let's solve the linear programming problem using the **method of corners (vertices)**. ### Problem Statement Minimize \(C = 6x + 7y\) Subject to: \[ \begin{align*} 4x + y &\geq 42 \\ 2x + y &\geq 30 \\ x + 3y &\geq 30 \\ x &\geq \\ y &\geq \\ \end{align*} \] --- ## Step 1: Find the Corner Points Let's convert the inequalities to equalities to find the intersection points (corners). ### 1. \(4x + y = 42\) ### 2. \(2x + y = 30\) ### 3. \(x + 3y = 30\) #### Intersection of (1) and (2): \[ \begin{align*} 4x + y &= 42 \\ 2x + y &= 30 \\ \end{align*} \] Subtract (2) from (1): \[ (4x + y) - (2x + y) = 42 - 30 \implies 2x = 12 \implies x = 6 \] Plug \(x = 6\) into (2): \[ 2(6) + y = 30 \implies 12 + y = 30 \implies y = 18 \] So, point A: **(6, 18)** --- #### Intersection of (1) and (3): \[ \begin{align*} 4x + y &= 42 \\ x + 3y &= 30 \\ \end{align*} \] From (3): \(x = 30 - 3y\) Plug into (1): \[ 4(30 - 3y) + y = 42 \implies 120 - 12y + y = 42 \implies 120 - 11y = 42 \implies 11y = 78 \implies y = \frac{78}{11} \] Plug \(y = \frac{78}{11}\) into \(x = 30 - 3y\): \[ x = 30 - 3 \times \frac{78}{11} = 30 - \frac{234}{11} = \frac{330 - 234}{11} = \frac{96}{11} \] So, point B: \(\left(\frac{96}{11}, \frac{78}{11}\right)\) --- #### Intersection of (2) and (3): \[ \begin{align*} 2x + y &= 30 \\ x + 3y &= 30 \\ \end{align*} \] From (2): \(y = 30 - 2x\) Plug into (3): \[ x + 3(30 - 2x) = 30 \implies x + 90 - 6x = 30 \implies -5x = -60 \implies x = 12 \] Plug \(x = 12\) into \(y = 30 - 2x\): \[ y = 30 - 2 \times 12 = 6 \] So, point C: **(12, 6)** --- ## Step 2: Verify Feasibility All constraints are "\(\geq\)" with \(x, y \geq \), so the feasible region is **above** all the lines in the first quadrant. Check that each point satisfies all constraints: - **(6, 18):** - \(4(6) + 18 = 24 + 18 = 42 \geq 42\) - \(2(6) + 18 = 12 + 18 = 30 \geq 30\) - \(6 + 3(18) = 6 + 54 = 60 \geq 30\) - OK - **(12, 6):** - \(4(12) + 6 = 48 + 6 = 54 \geq 42\) - \(2(12) + 6 = 24 + 6 = 30 \geq 30\) - \(12 + 3(6) = 12 + 18 = 30 \geq 30\) - OK - **\(\left(\frac{96}{11}, \frac{78}{11}\right)\):** - Check feasibility: - \(4x + y = 4 \times \frac{96}{11} + \frac{78}{11} = \frac{384 + 78}{11} = \frac{462}{11} = 42\) - \(2x + y = 2 \times \frac{96}{11} + \frac{78}{11} = \frac{192 + 78}{11} = \frac{270}{11} \approx 24.545\) (**Not feasible**; must be \(\geq 30\)) - Therefore, this point is **not feasible**. ## Step 3: Evaluate Objective Function at Feasible Points \[ C = 6x + 7y \] - At (6, 18): \(C = 6 \times 6 + 7 \times 18 = 36 + 126 = 162\) - At (12, 6): \(C = 6 \times 12 + 7 \times 6 = 72 + 42 = 114\) ## Step 4: State the Minimum ### **Minimum value is \(C = 114\) at \((x, y) = (12, 6)\).** --- ## **Final Answer** ### The minimum value of \(C\) is: \[ \boxed{114} \] at \[ (x, y) = (12, 6) \] --- **Summary Table:** | Corner | \(C = 6x + 7y\) | |---------------|----------------| | (6, 18) | 162 | | (12, 6) | 114 | **Minimum at:** \((12, 6)\) with \(C = 114\)