Let's solve the problem **step by step** both graphically and using the **simplex method**. --- ### **Given Problem** Maximize and minimize \( z = x_1 + 2x_2 \) Subject to: \[ \begin{align*} x_1 + x_2 & \leq 800 \\ 0.125x_1 + 0.375x_2 & \leq 225 \\ 2x_1 + x_2 & \leq 1200 \\ x_1 & \geq 0 \\ x_2 & \geq 0 \end{align*} \] --- ## **a) Graphical Solution** ### **Step 1: Draw the Constraints** Let's rewrite constraints: 1. \(x_1 + x_2 \leq 800\) 2. \(0.125x_1 + 0.375x_2 \leq 225\) Divide both sides by 0.125: \(x_1 + 3x_2 \leq 1800\) 3. \(2x_1 + x_2 \leq 1200\) 4. \(x_1 \geq 0\) 5. \(x_2 \geq 0\) --- ### **Step 2: Find Intersections (Corner Points)** #### **Find intersection points of constraints:** #### **(i) \(x_1 + x_2 = 800\) and \(x_1 + 3x_2 = 1800\):** \[ x_1 + x_2 = 800 \implies x_1 = 800 - x_2 \] Substitute into second: \[ (800 - x_2) + 3x_2 = 1800 \\ 800 + 2x_2 = 1800 \\ 2x_2 = 1000 \implies x_2 = 500 \\ x_1 = 800 - 500 = 300 \] So, \(\boxed{(300, 500)}\) --- #### **(ii) \(x_1 + x_2 = 800\) and \(2x_1 + x_2 = 1200\):** \[ x_1 + x_2 = 800 \implies x_1 = 800 - x_2 \\ 2x_1 + x_2 = 1200 \\ 2(800 - x_2) + x_2 = 1200 \\ 1600 - 2x_2 + x_2 = 1200 \\ 1600 - x_2 = 1200 \implies x_2 = 400 \\ x_1 = 800 - 400 = 400 \] So, \(\boxed{(400, 400)}\) --- #### **(iii) \(x_1 + 3x_2 = 1800\) and \(2x_1 + x_2 = 1200\):** \[ x_1 + 3x_2 = 1800 \implies x_1 = 1800 - 3x_2 \\ 2x_1 + x_2 = 1200 \\ 2(1800 - 3x_2) + x_2 = 1200 \\ 3600 - 6x_2 + x_2 = 1200 \\ 3600 - 5x_2 = 1200 \implies 5x_2 = 2400 \implies x_2 = 480 \\ x_1 = 1800 - 3(480) = 1800 - 1440 = 360 \] So, \(\boxed{(360, 480)}\) --- #### **Intersections with axes:** ##### For \(x_1 + x_2 = 800\): - \(x_1=0\): \(x_2=800\) \(\to (0,800)\) - \(x_2=0\): \(x_1=800\) \(\to (800,0)\) ##### For \(x_1 + 3x_2 = 1800\): - \(x_1=0\): \(3x_2=1800 \implies x_2=600\) \(\to (0,600)\) - \(x_2=0\): \(x_1=1800\) (but \(x_1=1800\) is outside other constraints) ##### For \(2x_1 + x_2 = 1200\): - \(x_1=0\): \(x_2=1200\) \(\to (0,1200)\) - \(x_2=0\): \(2x_1=1200 \implies x_1=600\) \(\to (600,0)\) --- #### **Now check which of these are within all constraints:** Let’s check the feasible region bounded by: - \(x_1 + x_2 \leq 800\) - \(x_1 + 3x_2 \leq 1800\) - \(2x_1 + x_2 \leq 1200\) - \(x_1 \geq 0\) - \(x_2 \geq 0\) The feasible region is the intersection of all these inequalities. The valid corner points (from above and axes) are: - (0,0) - (0,600): \(x_1+3(600)=1800\), \(x_1=0\) OK; \(0+600=600\leq800\), \(2(0)+600=600\leq1200\) - (300,500): checked above - (360,480): checked above - (400,400): checked above - (600,0): \(2(600)=1200\), \(600+0=600\leq800\), \(600+3(0)=600\leq1800\) - (800,0): \(800+0=800\leq800\), \(800+3(0)=800\leq1800\), \(2(800)=1600>1200\) **not feasible** --- ### **Step 3: Evaluate Objective Function at Each Vertex** \[ z = x_1 + 2x_2 \] - At **(0,0):** \(z=0+2\cdot0=0\) - At **(0,600):** \(z=0+2\cdot600=1200\) - At **(300,500):** \(z=300+2\cdot500=300+1000=1300\) - At **(360,480):** \(z=360+2\cdot480=360+960=1320\) - At **(400,400):** \(z=400+2\cdot400=400+800=1200\) - At **(600,0):** \(z=600+2\cdot0=600\) --- ### **Step 4: Find Maximum and Minimum** - **Maximum:** at (360,480), \(z=1320\) - **Minimum:** at (0,0), \(z=0\) --- ### **Graphical Conclusion** - **Maximum \(z=1320\) at \((360,480)\)** - **Minimum \(z=0\) at \((0,0)\)** --- ## **b) Simplex Method Solution** Convert inequalities to equalities by adding slack variables: \[ \begin{align*} x_1 + x_2 + s_1 &= 800 \\ x_1 + 3x_2 + s_2 &= 1800 \\ 2x_1 + x_2 + s_3 &= 1200 \\ x_1, x_2, s_1, s_2, s_3 &\geq 0 \end{align*} \] Objective function: Maximize \(z = x_1 + 2x_2\) Set up the simplex tableau: | BV | \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | \(s_3\) | RHS | |-----|:------:|:------:|:------:|:------:|:------:|:----:| | \(s_1\) | 1 | 1 | 1 | 0 | 0 | 800 | | \(s_2\) | 1 | 3 | 0 | 1 | 0 | 1800 | | \(s_3\) | 2 | 1 | 0 | 0 | 1 | 1200 | | \(z\) | -1 | -2 | 0 | 0 | 0 | 0 | **Proceed with the simplex method** (this will be a bit long, but here’s the process): #### **Step 1: Identify entering variable (most negative in z-row): \(x_2\) (-2)** #### **Step 2: Ratio test** - Row 1: \(800/1 = 800\) - Row 2: \(1800/3 = 600\) - Row 3: \(1200/1 = 1200\) Smallest is 600 (Row 2), so \(x_2\) enters, \(s_2\) leaves. #### **Step 3: Pivot on Row 2, Col \(x_2\) (make it 1, others 0 in \(x_2\) column)** - Divide Row 2 by 3: - \(x_1 + 3x_2 + s_2 = 1800\) - \(x_1/3 + x_2 + s_2/3 = 600\) New tableau: | BV | \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | \(s_3\) | RHS | |-----|:------:|:------:|:------:|:------:|:------:|:----:| | \(s_1\) | 1 | 1 | 1 | 0 | 0 | 800 | | \(x_2\) | 1/3 | 1 | 0 | 1/3 | 0 | 600 | | \(s_3\) | 2 | 1 | 0 | 0 | 1 | 1200 | | \(z\) | -1 | -2 | 0 | 0 | 0 | 0 | Now eliminate \(x_2\) from other rows: - Row 1: Row 1 - Row 2 - Row 3: Row 3 - Row 2 - \(z\) row: \(z + 2\times\) Row 2 Let’s do the calculations: **Row 1:** \(s_1: 1 - (1/3) = 2/3\) for \(x_1\) \(1 - 1 = 0\) for \(x_2\) \(1 - 0 = 1\) for \(s_1\) \(0 - (1/3) = -1/3\) for \(s_2\) \(0 - 0 = 0\) for \(s_3\) \(800 - 600 = 200\) RHS **Row 3:** \(2 - (1/3 \times 1) = 2 - 1/3 = 5/3\) for \(x_1\) \(1 - 1 = 0\) for \(x_2\) \(0 - 0 = 0\) for \(s_1\) \(0 - (1/3) = -1/3\) for \(s_2\) \(1 - 0 = 1\) for \(s_3\) \(1200 - 600 = 600\) RHS **z row:** \(-1 + 2 \times (1/3) = -1 + 2/3 = -1/3\) for \(x_1\) \(-2 + 2 \times 1 = 0\) for \(x_2\) \(0 + 2 \times 0 = 0\) for \(s_1\) \(0 + 2 \times (1/3) = 2/3\) for \(s_2\) \(0 + 2 \times 0 = 0\) for \(s_3\) \(0 + 2 \times 600 = 1200\) RHS Tableau becomes: | BV | \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | \(s_3\) | RHS | |-----|:------:|:------:|:------:|:------:|:------:|:----:| | \(s_1\) | 2/3 | 0 | 1 | -1/3 | 0 | 200 | | \(x_2\) | 1/3 | 1 | 0 | 1/3 | 0 | 600 | | \(s_3\) | 5/3 | 0 | 0 | -1/3 | 1 | 600 | | \(z\) | -1/3 | 0