1. The problem involves deciding how many 100-ounce lots of plastic Parket Sisters should buy at $6.00 per ounce (usual cost $5.00 plus $1.00) and determining the optimal product mix for maximized profit. 2. Given constraints: - Plastic limit: 1,000 ounces - Chrome limit: 1,200 ounces - Stainless steel limit: 2,000 ounces 3. Product variables: $X_1$ = ballpoint pens, $X_2$ = mechanical pencils, $X_3$ = fountain pens. 4. Each unit requires plastic, chrome, stainless steel in amounts: - Ballpoint: 1.2, 0.8, 2.0 ounces respectively - Mechanical pencil: 1.7, 0, 3.0 ounces respectively - Fountain pen: 1.2, 2.3, 4.5 ounces respectively 5. Profit per item: $3.00X_1 + 3.00X_2 + 5.00X_3$ 6. Lot size: 100 ounces of plastic per lot. Parket Sisters must decide number of lots to buy, $L$, where $L \times 100 \leq 1000$ (max plastic available) and plastic cost $6.00$ per ounce. 7. Set plastic supply constraint: $1.2X_1 + 1.7X_2 + 1.2X_3 \leq 100L$ 8. Other constraints remain: $0.8X_1 + 0X_2 + 2.3X_3 \leq 1200$ $2.0X_1 + 3.0X_2 + 4.5X_3 \leq 2000$ $X_1, X_2, X_3, L \geq 0$ 9. To find optimal lots $L$: try each integer $L$ from 1 to 10 (since $10 \times 100 = 1000$ ounces plastic max). For each $L$, solve the linear program for $(X_1,X_2,X_3)$ to maximize profit. 10. The best $L$ value maximizes total profit minus additional plastic cost: total plastic cost = $6 imes 100 imes L = 600L$. 11. Solve LP for each $L$ to find product mix maximizing profit. Since the cost per ounce is fixed over usual, focus on max total profit. 12. This problem requires computational LP solving for various $L$; generally, buy maximum lots $L=10$ if profit increase offsets cost. **Final:** Buy up to 10 lots (1000 ounces plastic), maximize $3X_1 + 3X_2 + 5X_3$ subject to constraints with adjusted plastic supply $\leq 1000$ ounces. Optimal product mix depends on LP solution for given $L$, and optimal profit is maximum objective function value at that solution.