Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.3

Question 1. The following table gives the annual demand and unit price of 3 items

Ordering cost is Rs. 5 per order and holding cost is 10% of unit price. Determine the following:
(i) EOQ in units
(ii) Minimum average cost
(iii) EOQ in rupees
(iv) EOQ in years of supply
(v) Number of orders per year
Answer:
**Item A:**
Demand rate, \( R = 800 \)
Ordering cost, \( C_3 = \text{Rs. } 5 \)
Carrying cost \( C_1 = 10\% \) of unit price
\( = \frac{10}{100} \times 0.02 = 0.002 \)
(i) EOQ in units:
\[ \text{EOQ} = \sqrt{\frac{2RC_3}{C_1}} \] \[ = \sqrt{\frac{2 \times 800 \times 5}{0.002}} \] To remove the decimal from the denominator, multiply numerator and denominator by 1000.
\[ = \sqrt{\frac{2 \times 800 \times 5 \times 1000}{0.002 \times 1000}} \] \[ = \sqrt{\frac{800 \times 100 \times 100}{2}} \] \[ = \sqrt{400 \times 100 \times 100} \] \[ = 20 \times 10 \times 10 \] \[ = 2000 \text{ units} \] (ii) Minimum Average Cost = \( C_0 \)
\[ C_0 = \sqrt{2RC_3C_1} \] \[ = \sqrt{2 \times 800 \times 5 \times 0.002} \] \[ = \sqrt{800 \times 0.02} \] \[ = \sqrt{16.00} \] \[ = 4 \text{ Rs.} \] (iii) EOQ in rupees = EOQ \( \times \) Unit price
\[ = 2000 \times 0.02 \] \[ = 2000 \times \frac{2}{100} \] \[ = \text{Rs. } 40 \] (iv) EOQ in years of supply = \( \frac{\text{EOQ}}{\text{Demand}} \)
\[ = \frac{2000}{800} = 2.5 \] (v) Number of orders per year = \( \frac{\text{Demand}}{\text{EOQ}} \)
\[ = \frac{800}{2000} = 0.4 \]
**Item B:**
Demand rate, \( R = 400 \)
Ordering cost, \( C_3 = \text{Rs. } 5 \)
Carrying cost \( C_1 = 10\% \) of \( 1.00 \)
\( = \frac{10}{100} \times 1.00 = 0.1 \)
(i) EOQ in units:
\[ \text{EOQ} = \sqrt{\frac{2RC_3}{C_1}} \] \[ = \sqrt{\frac{2 \times 400 \times 5}{0.1}} \] To remove the decimal from the denominator, multiply numerator and denominator by 10.
\[ = \sqrt{\frac{2 \times 400 \times 5 \times 10}{0.1 \times 10}} \] \[ = \sqrt{\frac{2 \times 400 \times 5 \times 10}{1}} \] \[ = \sqrt{40000} \] \[ = 200 \text{ units} \] (ii) Minimum Average Cost = \( C_0 \)
\[ C_0 = \sqrt{2RC_3C_1} \] \[ = \sqrt{2 \times 400 \times 5 \times 0.1} \] \[ = \sqrt{400} \] \[ = \text{Rs. } 20 \] (iii) EOQ in rupees = EOQ \( \times \) Unit price
\[ = 200 \times 1 \] \[ = \text{Rs. } 200 \] (iv) EOQ in years of supply = \( \frac{\text{EOQ}}{\text{Demand}} \)
\[ = \frac{200}{400} = 0.5 \] (v) Number of orders per year = \( \frac{\text{Demand}}{\text{EOQ}} \)
\[ = \frac{400}{200} = 2 \]
**Item C:**
Annual Demand, \( R = 13800 \)
Ordering cost, \( C_3 = \text{Rs. } 5 \)
Carrying cost, \( C_1 = 10\% \) of unit price
\( = \frac{10}{100} \times 0.20 = 0.02 \)
(i) EOQ in units:
\[ \text{EOQ} = \sqrt{\frac{2RC_3}{C_1}} \] \[ = \sqrt{\frac{2 \times 13800 \times 5}{0.02}} \] To remove the decimal from the denominator, multiply numerator and denominator by 100.
\[ = \sqrt{\frac{2 \times 13800 \times 5 \times 100}{0.02 \times 100}} \] \[ = \sqrt{\frac{2 \times 13800 \times 5 \times 100}{2}} \] \[ = \sqrt{138 \times 100 \times 5 \times 100} \] \[ = \sqrt{6900000} \] \[ = 10 \times 10 \times \sqrt{690} \] \[ = 100 \times 26.2678 \] \[ = 2627 \text{ units (approx)} \] (ii) Minimum Average Cost = \( C_0 \)
\[ C_0 = \sqrt{2RC_3C_1} \] \[ = \sqrt{2 \times 13800 \times 5 \times 0.02} \] \[ = \sqrt{2760} \] \[ = 52.535 \] \[ = \text{Rs. } 52.54 \text{ (approx)} \] (iii) EOQ in rupees = EOQ \( \times \) Unit price
\[ = 2627 \times 0.20 \] \[ = \text{Rs. } 525.40 \] (iv) EOQ in years of supply = \( \frac{\text{EOQ}}{\text{Demand}} \)
\[ = \frac{2627}{13800} = 0.19036 \approx 0.19 \] (v) Number of orders per year = \( \frac{\text{Demand}}{\text{EOQ}} \)
\[ = \frac{13800}{2627} = 5.2531 \approx 5.25 \]In simple words: We calculated the best quantity to order for each item to reduce total costs. This is called the Economic Order Quantity (EOQ). We also found the lowest possible total cost and how many times we need to order each year.

🎯 Exam Tip: Remember to use the correct values for demand, ordering cost, and carrying cost for each item when calculating EOQ and minimum average cost. Pay attention to unit consistency (e.g., annual demand, annual carrying cost).

Question 2. A dealer has to supply his customer with 400 units of a product per week. The dealer gets the product from the manufacturer at a cost of Rs. 50 per unit. The cost of ordering from the manufacturers in Rs. 75 per order. The cost of holding inventory is 7.5 % per year of the product cost. Find
(i) EOQ
(ii) Total optimum cost
Answer:
Demand = 400 units per week
Annual demand \( R = 400 \times 52 = 20800 \) units per year
Ordering cost per order \( C_3 = \text{Rs. } 75 \)
Inventory cost \( C_1 = 7.5\% \) per year of the cost
\[ = \frac{7.5}{100} \times 50 \text{ per year} \] \[ = 0.075 \times 50 = \text{Rs. } 3.75 \text{ per year} \] (i) EOQ in units:
\[ \text{EOQ} = \sqrt{\frac{2RC_3}{C_1}} \] \[ = \sqrt{\frac{2 \times 20800 \times 75}{3.75}} \] To remove the decimal from the denominator, multiply numerator and denominator by 100.
\[ = \sqrt{\frac{2 \times 20800 \times 75 \times 100}{3.75 \times 100}} \] \[ = \sqrt{\frac{2 \times 20800 \times 75 \times 100}{375}} \] \[ = \sqrt{832000} \] \[ \text{EOQ} = 912.14 \approx 912 \text{ units} \] (ii) Total optimum cost = Purchasing cost \( + \) Minimum annual cost
\[ = (R \times \text{Unit Price}) + \sqrt{2RC_3C_1} \] \[ = (20800 \times 50) + \sqrt{2 \times 20800 \times 75 \times 3.75} \] \[ = 1040000 + \sqrt{11700000} \] \[ = 1040000 + 3420.526 \] \[ = \text{Rs. } 1,043,420.53 \text{ per year} \]In simple words: First, we calculated the yearly demand for the product. Then, using the ordering and holding costs, we found the best quantity to order at a time (EOQ) to keep costs down. Finally, we calculated the lowest possible total cost for buying and storing the product over a year.

🎯 Exam Tip: Always make sure to convert all costs and demand rates to a consistent time period (usually annual) before applying the EOQ formula. Pay attention to whether the holding cost is a percentage of the unit price or a fixed amount.