Question: A fashion company manufactures two models of short pants, Short pant A and Short pant B. Each short pants is first cut in the trimming shop and then sent to the finishing shop where it is stitched, button-holed, and packed. The number of man-hours of labour required in each shop per 100 shorts is given below: Shop Trimming Shop Short pant A 1 Short pant B 2 Finishing Shop 4 9 Because of limitations in plant capacity, no more than 400 man-hours are available in the Trimming Shop and 6000 man-hours are available in the Finishing Shop during the next six months. The contribution from sales is Rs. 12 per Short pant A and Rs. 20 per Short pant B. Assume that there is no shortage of raw materials and that market demand is unlimited. Determine the optimal product mix using the graphical method.

1. **State the problem:** We want to determine the optimal number of Short pant A and Short pant B to produce to maximize profit, given labor hour constraints in the trimming and finishing shops. 2. **Define variables:** Let $x$ = number of Short pant A produced (in hundreds) Let $y$ = number of Short pant B produced (in hundreds) 3. **Formulate the objective function:** Profit per 100 units of A = $12 \times 100 = 1200$ (but since $x,y$ are in hundreds, we use 12 and 20 directly) Profit per 100 units of B = $20 \times 100 = 2000$ Objective: Maximize $$Z = 12x + 20y$$ 4. **Formulate constraints based on labor hours:** - Trimming shop constraint: $$1x + 2y \leq 400$$ - Finishing shop constraint: $$4x + 9y \leq 6000$$ - Non-negativity constraints: $$x \geq 0, \quad y \geq 0$$ 5. **Graphical method steps:** - Plot the lines: $$1x + 2y = 400$$ $$4x + 9y = 6000$$ - Find intercepts: For trimming shop: - If $x=0$, then $2y=400 \Rightarrow y=200$ - If $y=0$, then $x=400$ For finishing shop: - If $x=0$, then $9y=6000 \Rightarrow y=\frac{6000}{9} = 666.67$ - If $y=0$, then $4x=6000 \Rightarrow x=1500$ 6. **Find corner points of feasible region:** - Intersection of trimming and finishing constraints: Solve system: $$\begin{cases} x + 2y = 400 \\ 4x + 9y = 6000 \end{cases}$$ Multiply first equation by 4: $$4x + 8y = 1600$$ Subtract from second: $$4x + 9y - (4x + 8y) = 6000 - 1600 \Rightarrow y = 440$$ Substitute $y=440$ into first: $$x + 2(440) = 400 \Rightarrow x + 880 = 400 \Rightarrow x = 400 - 880 = -480$$ Since $x$ cannot be negative, this intersection is outside feasible region. 7. **Check feasible corner points:** - At $x=0$, $y$ limited by trimming: $2y \leq 400 \Rightarrow y \leq 200$ - At $x=0$, $y$ limited by finishing: $9y \leq 6000 \Rightarrow y \leq 666.67$ So $y \leq 200$ at $x=0$. - At $y=0$, $x$ limited by trimming: $x \leq 400$ - At $y=0$, $x$ limited by finishing: $4x \leq 6000 \Rightarrow x \leq 1500$ So $x \leq 400$ at $y=0$. 8. **Evaluate objective function at corner points:** - Point A: $(0,0)$ $$Z = 12(0) + 20(0) = 0$$ - Point B: $(0,200)$ $$Z = 12(0) + 20(200) = 4000$$ - Point C: $(400,0)$ $$Z = 12(400) + 20(0) = 4800$$ - Point D: Intersection of finishing constraint with trimming constraint boundary at $x=400$: Check if $x=400$ satisfies finishing constraint: $$4(400) + 9y \leq 6000 \Rightarrow 1600 + 9y \leq 6000 \Rightarrow 9y \leq 4400 \Rightarrow y \leq 488.89$$ At $x=400$, max $y=0$ (since trimming constraint is $x + 2y \leq 400$ and $400 + 2y \leq 400 \Rightarrow y \leq 0$), so no positive $y$ here. - Point E: Intersection of finishing constraint with $y=0$ is $(1500,0)$ but trimming constraint limits $x \leq 400$, so not feasible. 9. **Conclusion:** The feasible region is bounded by points $(0,0)$, $(0,200)$, and $(400,0)$. Maximum profit is at $(400,0)$ with $$Z = 4800$$ 10. **Answer:** Produce 400 units of Short pant A and 0 units of Short pant B to maximize profit under given constraints.