Q.64 The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quantity produced. The fixed cost (F) of production of the same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V + F)?
Minimize Total Cost: Variable Cost 4q + Fixed Cost 100/q
Answer: To minimize total cost where variable cost V = 4q and fixed cost F = 100/q, the optimal quantity is 5 units. This balances the increasing variable costs against decreasing fixed costs per unit.
Total Cost Function
Total cost TC = V + F = 4q + 100/q. To find the minimum, take the first derivative: d(TC)/dq = 4 - 100/q².
Set d(TC)/dq = 0: 4 = 100/q², so q² = 25, and q = 5 (since quantity is positive).
The second derivative d²(TC)/dq² = 200/q³ is positive at q=5, confirming a minimum.
Option Analysis
Calculate TC for each integer option near 5:
| Quantity (q) | Variable Cost (4q) | Fixed Cost (100/q) | Total Cost (TC) |
|---|---|---|---|
| 4 | 16 | 25 | 41[web:9] |
| 5 | 20 | 20 | 40 [web:9] |
| 6 | 24 | 16.67 | 40.67[web:9] |
| 7 | 28 | 14.29 | 42.29[web:9] |
Option (A) 5 yields the lowest TC at 40. Options (B) 4, (D) 6, and (C) 7 all give higher costs.
Cost Comparison Summary
| q (Units) | V=4q | F=100/q | TC | Status |
|---|---|---|---|---|
| 4 | 16 | 25 | 41 | Higher |
| 5 | 20 | 20 | 40 | ✅ Minimum |
| 6 | 24 | 16.67 | 40.67 | Higher |
| 7 | 28 | 14.29 | 42.29 | Higher |
This table shows 5 units as optimal for minimize total cost V=4q F=100/q.
Applications in Biochemical Engineering
In bioprocesses like enzyme production, variable costs (media, labor) scale with q, while fixed costs (setup, equipment amortization) decline per unit. Optimizing at q=5 minimizes expenses, aiding scalability in microbial fermentation or bioreactor runs.
🎯 Final Answer
(A) 5 units – This is the quantity that minimizes total production cost.
