1. **State the problem:** We have 3 items with weights $w_1=10$, $w_2=20$, $w_3=30$ and values $v_1=60$, $v_2=100$, $v_3=120$. We want to maximize the total value in a knapsack with capacity 40 units by choosing weights $x_1, x_2, x_3$ of each item such that $x_1+x_2+x_3 \leq 40$ and $0 \leq x_i \leq w_i$. 2. **Formula and approach:** The fractional knapsack problem is solved by taking items in order of their value-to-weight ratio $\frac{v_i}{w_i}$ from highest to lowest until the capacity is full. Calculate ratios: $$\frac{v_1}{w_1} = \frac{60}{10} = 6, \quad \frac{v_2}{w_2} = \frac{100}{20} = 5, \quad \frac{v_3}{w_3} = \frac{120}{30} = 4.$$ 3. **Select items by ratio:** Start with item 1 (ratio 6), then item 2 (ratio 5), then item 3 (ratio 4). 4. **Fill knapsack:** - Take full $x_1 = 10$ units of item 1. - Remaining capacity: $40 - 10 = 30$. - Take full $x_2 = 20$ units of item 2. - Remaining capacity: $30 - 20 = 10$. - Take $x_3 = 10$ units of item 3 (partial, since only 10 units left). 5. **Check total weight:** $$x_1 + x_2 + x_3 = 10 + 20 + 10 = 40,$$ which fits the capacity. 6. **Calculate total value:** $$V = 6 \times 10 + 5 \times 20 + 4 \times 10 = 60 + 100 + 40 = 200.$$ **Final answer:** Take $x_1=10$, $x_2=20$, and $x_3=10$ units of items 1, 2, and 3 respectively to maximize value 200.