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## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 6 Applications of Differentiation Ex 6.3

### Samacheer Kalvi 11th Business Maths Applications of Differentiation Ex 6.3 Text Book Back Questions and Answers

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

Ordering cost is ₹ 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
Solution:
Item A:
Demand rate, R = 800
Ordering cost, C3 = ₹ 5
Carrying cost C1 = 10% of unit price
= $$\frac{10}{100}$$ × 0.02

(i) EOQ in units

(ii) Minimum Average Cost = C0 = $$\sqrt{2 \mathrm{RC}_{3} \mathrm{C}_{1}}$$
= $$\sqrt{2 \times 800 \times 5 \times \frac{10}{100} \times 0.02}$$
= $$\sqrt{800 \times 0.02}$$
= $$\sqrt{16.00}$$
= ₹ 4

(iii) EOQ in rupees = EOQ × Unit price
= 2000 × 0.02
= 2000 × $$\frac{2}{100}$$
= ₹ 40

(iv) $$\frac{\mathrm{EOQ}}{\text { Demand }}=\frac{2000}{800}$$ = 2.5

(v) $$\frac{\text { Demand }}{\mathrm{EOQ}}=\frac{800}{2000}$$ = 0.4

Item B:
Demand rate, R = 400
Ordering cost, C3 = ₹ 5
Carrying cost C1 = 10% of 1.00

(i) EOQ in units

= 20 × 10
= 200 units

(ii) Minimum Average Cost = C0 = $$\sqrt{2 \mathrm{RC}_{3} \mathrm{C}_{1}}$$
= $$\sqrt{2 \times 400 \times 5 \times \frac{10}{100} \times 1}$$
= √400
= ₹ 20

(iii) EOQ in rupees = EOQ × unit price
= 200 × 1
= ₹ 200

(iv) $$\frac{\mathrm{EOQ}}{\text { Demand }}=\frac{200}{400}$$ = 0.5

(v) $$\frac{\text { Demand }}{\mathrm{EOQ}}=\frac{400}{200}$$ = 2

Item C:
Annual Demand, R = 800
Ordering cost, C3 = ₹ 5
Carrying cost, C1 = 10% of unit price
= $$\frac{10}{100}$$ × 0.20
= $$\frac{2}{100}$$

(i) EOQ in units

= 100 × 26.2678
= 100 × 26.27
= 2627

(ii) Minimum Average Cost = C0 = $$\sqrt{2 \mathrm{RC}_{3} \mathrm{C}_{1}}$$
= $$\sqrt{2 \times 13800 \times 5 \times \frac{10}{100} \times 0.2}$$
= $$\sqrt{2 \times 138 \times 5 \times 2}$$
= $$\sqrt{2760}$$
= 52.535
= ₹ 52.54

(iii) EOQ in rupees = 2627 × 0.20 = ₹ 25.40 [∵ Unit price = 0.20]

(iv) $$\frac{\mathrm{EOQ}}{\text { Demand }}=\frac{2627}{13800}$$ = 0.19036 = 0.19

(v) $$\frac{\text { Demand }}{\mathrm{EOQ}}=\frac{13800}{2627}$$ = 5.2531 = 5.25

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 ₹ 50 per unit. The cost of ordering from the manufacturers in ₹ 75 per order. The cost of holding inventory is 7.5 % per year of the product cost. Find
(i) EOQ
(ii) Total optimum cost.
Solution:
Demand = 400 units per week
Annual demand = 400 × 52 per year
Ordering cost per order C3 = 175
Inventory cost C1 = 7.5% per year of the cost
= 7.5% of 50 per year
= $$\frac{7.5}{100}$$ × 50
= $$\frac{7.5 \times 50}{100 \times 52}$$ (per week)
(i) EOQ in units

EOQ = 912 units

(ii) Total optimum cost = Purchasing cost + Minimum annual cost

= 20000 + $$\sqrt{\frac{225000}{52}}$$
= 20000 + √4326.92307
= 20000 + √4326.9231
= 20000 + 65.7793
= ₹ 20,065.78 per week