Strategic Cost Management · Business Application of Maxima and Minima

Inventory Control and Economic Order Quantity Using Maxima and Minima

Updated 11 October 2026 · Fact-checked

Economic Order Quantity (EOQ) is the order size that minimises total yearly inventory cost, which is ordering cost plus carrying cost. Write total cost as a function of order size Q, set dTC/dQ = 0 to get Q = √(2AO ÷ C), then confirm with the second derivative that it is a minimum.

Understand Inventory Control and Economic Order Quantity

Every business that holds stock faces a trade-off. If you order in small lots, you place many orders and pay ordering cost again and again. If you order in large lots, you place few orders but hold more stock, so carrying cost rises.

Total inventory cost is the sum of these two. Ordering cost falls as order size rises. Carrying cost rises as order size rises. A sum of a falling curve and a rising curve has a lowest point. Calculus finds that point exactly.

Let Q be the order size, A the annual demand, O the cost per order and C the carrying cost per unit per year. Number of orders is A ÷ Q. Average stock is Q ÷ 2, assuming demand is steady and stock falls evenly to zero before each new lot arrives. So TC = (A ÷ Q) × O + (Q ÷ 2) × C.

Differentiate with respect to Q and set it to zero. You get Q² = 2AO ÷ C. The second derivative is 2AO ÷ Q³, which is positive for Q > 0, so the point is a minimum. At EOQ, ordering cost equals carrying cost. This is a useful check on your answer.

The model assumes constant demand, a fixed ordering cost per order, a carrying cost proportional to units held, instant delivery and no stock-outs. Exam questions may add a purchase price, quantity discounts or a constraint. State your assumptions when the question is open.

Key rules to remember

Total inventory cost: TC(Q) = (A ÷ Q) × O + (Q ÷ 2) × C; Add the purchase cost A × P only if price varies with order size. Otherwise it does not affect EOQ.
First-order condition: dTC/dQ = −AO ÷ Q² + C ÷ 2 = 0; Setting this to zero gives the stationary point.
Economic Order Quantity: EOQ = √(2AO ÷ C); If carrying cost is a percentage i of unit price P, then C = i × P.
Second-order condition: d²TC/dQ² = 2AO ÷ Q³ > 0 for Q > 0; A positive value confirms a minimum.
Minimum total cost: TC at EOQ = √(2AOC); Excluding purchase cost. Ordering cost equals carrying cost at this point.
Number of orders and cycle: Orders = A ÷ EOQ; Cycle time = EOQ ÷ A (in years); Multiply cycle time by 365 or 300 as the question states, to get days.

How to solve Inventory Control and Economic Order Quantity questions

Use this method for any EOQ question that asks you to derive or apply the optimum order size using calculus.

1List A (annual demand), O (cost per order) and C (carrying cost per unit per year). Convert monthly or percentage figures to yearly per-unit values.
2Write the total cost function: TC = (A ÷ Q) × O + (Q ÷ 2) × C. Add purchase cost only if price depends on Q.
3Differentiate TC with respect to Q. Write dTC/dQ = −AO ÷ Q² + C ÷ 2.
4Set dTC/dQ = 0 and solve: Q² = 2AO ÷ C, so Q = √(2AO ÷ C). Reject the negative root.
5Find d²TC/dQ² = 2AO ÷ Q³. State that it is positive for Q > 0, so Q is a minimum.
6Compute the number of orders, the cycle time and the minimum total cost as asked.
7Check that ordering cost equals carrying cost at EOQ, then write the answer with units and a short recommendation.

Quickest way: Direct formula with a balance check

When to use it: Use it in MCQs, or in a written answer where the question says 'find EOQ' and does not ask for a derivation.

Identify A, O and C. Make sure C is per unit per year.
Calculate EOQ = √(2AO ÷ C) and simplify under the root before taking it.
Check: (A ÷ EOQ) × O should equal (EOQ ÷ 2) × C.
Minimum cost = 2 × ordering cost at EOQ, or √(2AOC).
If a derivation is asked, add two lines: dTC/dQ = 0 and d²TC/dQ² > 0.

Common mistakes in Inventory Control and Economic Order Quantity

Using total order cost or carrying cost for the wrong period, such as monthly carrying cost with yearly demand.
Data is given in mixed time units and students plug numbers straight into the formula.
Fix: Convert A and C to the same period, normally one year, before calculating.
Taking carrying cost as the average stock times C but using Q instead of Q ÷ 2.
Students forget that stock falls from Q to zero, so the average is half of Q.
Fix: Always write carrying cost as (Q ÷ 2) × C when building the function.
Skipping the second-order test.
Students remember the formula and treat derivation as a formality.
Fix: Write d²TC/dQ² = 2AO ÷ Q³ > 0 for Q > 0. It earns a mark in derivation questions.
Including the purchase price in carrying cost and also adding purchase cost to EOQ analysis.
Carrying cost given as a percentage of price is confused with the price itself.
Fix: Compute C = percentage × unit price. Purchase cost A × P is constant and drops out when differentiating, unless discounts apply.
Reporting EOQ as a decimal without rounding sensibly, or giving cost without the order count.
Students stop after the square root.
Fix: Round to a practical whole number only if asked, then state number of orders and total cost clearly.

Worked examples

Example 1

A Pune manufacturer uses 14,400 units of a component a year. Ordering cost is ₹200 per order. Carrying cost is ₹16 per unit per year. Using differentiation, find the EOQ, the number of orders a year and the minimum yearly ordering plus carrying cost.

Show the solution

A = 14,400; O = ₹200; C = ₹16.
TC = (14,400 ÷ Q) × 200 + (Q ÷ 2) × 16 = 28,80,000 ÷ Q + 8Q.
dTC/dQ = −28,80,000 ÷ Q² + 8 = 0, so Q² = 3,60,000.
Q = 600 units (negative root rejected).
d²TC/dQ² = 57,60,000 ÷ Q³, which is positive for Q > 0, so Q = 600 gives a minimum.
Orders per year = 14,400 ÷ 600 = 24.
Ordering cost = 24 × 200 = ₹4,800. Carrying cost = (600 ÷ 2) × 16 = ₹4,800. They are equal, as expected.
Minimum total cost = 4,800 + 4,800 = ₹9,600.

Answer: EOQ = 600 units; 24 orders a year; minimum ordering plus carrying cost = ₹9,600.

Example 2

Annual demand for a raw material is 5,000 units. Purchase price is ₹120 per unit. Carrying cost is 10% of purchase price per year. Cost per order is ₹300. Derive the EOQ and find the total yearly cost including purchase cost.

Show the solution

A = 5,000; O = ₹300; P = ₹120.
C = 10% × 120 = ₹12 per unit per year.
TC = (5,000 ÷ Q) × 300 + (Q ÷ 2) × 12 + 5,000 × 120 = 15,00,000 ÷ Q + 6Q + 6,00,000.
dTC/dQ = −15,00,000 ÷ Q² + 6 = 0, so Q² = 2,50,000.
Q = 500 units.
d²TC/dQ² = 30,00,000 ÷ Q³ > 0, so this is a minimum.
Ordering cost = (5,000 ÷ 500) × 300 = 10 × 300 = ₹3,000.
Carrying cost = (500 ÷ 2) × 12 = ₹3,000.
Total cost = 3,000 + 3,000 + 6,00,000 = ₹6,06,000.

Answer: EOQ = 500 units; total yearly cost including purchases = ₹6,06,000.

Exam tips

In MCQs, check units first. A wrong time base for carrying cost is the most common trap.
In descriptive answers, show the cost function, first derivative, EOQ and second derivative in that order. Marks follow these steps.
Use the equality of ordering and carrying cost at EOQ to verify your answer in seconds.
When carrying cost is a percentage of price, compute C in rupees first and write it down.
Close with a one-line recommendation, such as the number of orders and order size the firm should adopt.

Practice questions from Business Application of Maxima and Minima

Inventory Control and Economic Order Quantity in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Inventory Control and Economic Order Quantity: frequently asked questions

How do you derive EOQ using differentiation?

Write TC = (A ÷ Q) × O + (Q ÷ 2) × C. Differentiate with respect to Q and set the result to zero to get Q = √(2AO ÷ C). Then show the second derivative 2AO ÷ Q³ is positive, so it is a minimum.

Why does purchase cost not affect EOQ?

If the price per unit is constant, total purchase cost A × P does not change with order size. Its derivative with respect to Q is zero, so it drops out. It matters only when price changes with order size, as with quantity discounts.

Why is average inventory taken as Q ÷ 2?

The model assumes steady demand and instant replenishment. Stock starts at Q and falls evenly to zero, so the average held over the cycle is half of Q.

Do I need the second derivative in the exam?

Yes, if the question says derive or use calculus. It proves that the stationary point is a minimum, not a maximum. For a plain EOQ calculation, the formula is enough.