# Step-by-Step Solution Given the linear programming problem: **Minimize:** \( C =6x + 7y \**Subject to:** \[ begin{align*} 4x + y &\geq 42 \quad (1) \\ 2x + y &\geq 30 \ (2)\ x + 3y &\geq 30 \quad (3) \\ x \geq ,\ y \geq \end{align*} \] We solve by the **method of corners (vertices)**. --- ## 1. Convert inequalities for feasible region Since all constraints are "\(\geq\)", the feasible region is **above** each line. --- ## 2. Find intersection points ("corners") Let's find intersections of the constraints, as these are candidates for the minimum. ### Intersection A: (1) and (2) \[ \begin{align*} 4x + y &= 42 \\ 2x + y &= 30 \\ \end{align*} \] Subtract the second from the first: \[ (4x + y) - (2x + y) = 42 - 30 \implies 2x = 12 \implies x = 6 \] Substitute \(x=6\) into (2): \[ 2(6) + y = 30 \implies 12 + y = 30 \implies y = 18 \] **So, Intersection A:** \((6, 18)\) --- ### Intersection B: (1) and (3) \[ \begin{align*} 4x + y &= 42 \\ x + 3y &= 30 \\ \end{align*} \] Solve (3) for \(x\): \[ x = 30 - 3y \] Substitute into (1): \[ 4(30 - 3y) + y = 42 \\ 120 - 12y + y = 42 \\ 120 - 11y = 42 \\ -11y = 42 - 120 = -78 \\ y = \frac{-78}{-11} = \frac{78}{11} \] Now substitute \(y\) into \(x = 30 - 3y\): \[ x = 30 - 3\left(\frac{78}{11}\right) = 30 - \frac{234}{11} = \frac{330 - 234}{11} = \frac{96}{11} \] **So, Intersection B:** \(\left(\frac{96}{11}, \frac{78}{11}\right)\) --- ### Intersection C: (2) and (3) \[ \begin{align*} 2x + y &= 30 \\ x + 3y &= 30 \\ \end{align*} \] From (2): \(y = 30 - 2x\) Plug into (3): \[ x + 3(30 - 2x) = 30 \\ x + 90 - 6x = 30 \\ -5x + 90 = 30 \\ -5x = 30 - 90 = -60 \\ x = \frac{-60}{-5} = 12 \] Now \(y = 30 - 2x = 30 - 24 = 6\) **So, Intersection C:** \((12, 6)\) --- ## 3. Check if each point is in the feasible region - All constraints are "\(\geq\)" so the region is above all lines. - \(x \geq , y \geq \) is automatically satisfied by all found points. Let's check for each point: ### A: \((6, 18)\) - \(4x + y = 24 + 18 = 42 \geq 42\) ✔️ - \(2x + y = 12 + 18 = 30 \geq 30\) ✔️ - \(x + 3y = 6 + 54 = 60 \geq 30\) ✔️ ### B: \(\left(\frac{96}{11}, \frac{78}{11}\right)\) - \(4x + y = 4 \times \frac{96}{11} + \frac{78}{11} = \frac{384 + 78}{11} = \frac{462}{11} = 42\) ✔️ - \(2x + y = 2 \times \frac{96}{11} + \frac{78}{11} = \frac{192 + 78}{11} = \frac{270}{11} \approx 24.55\) But from the equations above, \(2x + y = 30\) (since that's the constraint used to find the intersection), so it should be \(30\), but let's check numerically: \[ 2 \times \frac{96}{11} + \frac{78}{11} = \frac{192 + 78}{11} = \frac{270}{11} = 24.545... \] But this is **less than 30**, so this point is actually **not feasible**. But this can't be, because the intersection is found by solving two equality lines. Let's check calculation: For (1) and (3): From above: - \(x = \frac{96}{11}\), \(y = \frac{78}{11}\) Now, check (2): \[ 2x + y = 2 \times \frac{96}{11} + \frac{78}{11} = \frac{192 + 78}{11} = \frac{270}{11} \approx 24.55 < 30 \] So, **this point is NOT in the feasible region**. --- ### C: \((12, 6)\) - \(4x + y = 48 + 6 = 54 \geq 42\) ✔️ - \(2x + y = 24 + 6 = 30 \geq 30\) ✔️ - \(x + 3y = 12 + 18 = 30 \geq 30\) ✔️ --- ## 4. Boundary points with axes? - \(x = \) or \(y = \) ### At \(x = \): - (1): \(y \geq 42\) - (2): \(y \geq 30\) - (3): \(3y \geq 30 \implies y \geq 10\) - So \(y \geq 42\), so point \((, 42)\) ### At \(y = \): - (1): \(4x \geq 42 \implies x \geq 10.5\) - (2): \(2x \geq 30 \implies x \geq 15\) - (3): \(x \geq 30\) - So, the strongest lower bound is \(x \geq 30\), so point \((30, )\) Now, check if these points satisfy all constraints: #### \((, 42)\): - (1): \(4 \times + 42 = 42 \geq 42\) ✔️ - (2): \(2 \times + 42 = 42 \geq 30\) ✔️ - (3): \( + 3 \times 42 = 126 \geq 30\) ✔️ #### \((30, )\): - (1): \(4 \times 30 + = 120 \geq 42\) ✔️ - (2): \(2 \times 30 + = 60 \geq 30\) ✔️ - (3): \(30 + = 30 \geq 30\) ✔️ --- ## 5. Calculate \(C\) at feasible corner points List of feasible corners: - \(A: (6, 18)\) - \(C: (12, 6)\) - \((, 42)\) - \((30, )\) Compute \(C\) at each: ### At \((6, 18)\): \[ C = 6 \times 6 + 7 \times 18 = 36 + 126 = 162 \] ### At \((12, 6)\): \[ C = 6 \times 12 + 7 \times 6 = 72 + 42 = 114 \] ### At \((, 42)\): \[ C = 6 \times + 7 \times 42 = + 294 = 294 \] ### At \((30, )\): \[ C = 6 \times 30 + 7 \times = 180 + = 180 \] --- ## 6. Find the minimum value Smallest \(C\) is at \((12, 6)\), where \(C = 114\). --- # Final Answer **The minimum value is:** \[ \boxed{C = 114 \quad \text{at} \quad (x, y) = (12, 6)} \] --- ## Summary Table | Point | \(C = 6x + 7y\) | |----------------|----------------| | (6, 18) | 162 | | (12, 6) | **114** | | (, 42) | 294 | | (30, ) | 180 | --- **Minimum is \(C = 114\) at \((x, y) = (12, 6)\).**