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Exercise 123Calculus › Single-variable calculus › Differentiation › Optimisation and curve sketchingL1Added 4 Nov 2026

The economic order quantity

#optimisation#economics#inventory#modelling

Statement

A shop sells D=1200D=1200 items per year at a steady rate. Each order costs K=50K=50 € whatever its size, and keeping one item in stock costs h=3h=3 € per year. If the shop orders qq items each time, the average stock is q2\frac q2. Find the order size qq that minimises the total annual cost.

Hint

The number of orders per year is Dq\frac Dq.

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Solution

Total cost.
C(q)=KDq⏟ordering+hq2⏟storage=60 000q+1.5 q,q>0.C(q)=\underbrace{\frac{KD}q}_{\text{ordering}}+\underbrace{\frac{hq}2}_{\text{storage}}=\frac{60\,000}q+1.5\,q,\qquad q>0.

Minimum. C′(q)=−60 000q2+1.5=0  ⟺  q2=40 000  ⟺  q=200C'(q)=-\frac{60\,000}{q^2}+1.5=0\iff q^2=40\,000\iff q=200. Since C→+∞C\to+\infty at 0+0^+ and at +∞+\infty and q=200q=200 is the only critical point, it is the global minimum (equivalently, C′<0C'<0 on (0,200)(0,200) and C′>0C'>0 on (200,∞)(200,\infty)).

Answer. Order 200200 items at a time (66 orders a year). The cost is C(200)=300+300=600C(200)=300+300=600 € per year.

General formula. q∗=2KDhq^*=\sqrt{\dfrac{2KD}h} (the Wilson formula). Here 2⋅50⋅1200/3=200\sqrt{2\cdot50\cdot1200/3}=200.

Remark. Again, ordering cost and storage cost are equal at the optimum. Treating qq as a real number is harmless here, because the optimum happens to be an integer.