Assume the company makes n order per year, then
Quantity per order is 100,000 / n
The maximum stock between orders is also 100,000 / n
So Cost per order = £35 + £7.00 ( (1/n) (100,000)/n )
Since there are n orders, the total cost per year is
C(n) = £35 n + £ 700,000 / n
Differentiating w.r.t. n,
C'(n) = 35 - 700,000 / n^2
Optimum value obtained at n such that C'(n) = 0
Therefore, 700,000 / n^2 = 35
==> n^2 = 700,000 / 35 = 20,000
==> n = sqrt(20,000) = sqrt(2) *100 = 141.4
===> n = 141
Therefore optimum quantity = 100,000/ 141 = 709 litres.
At this value of n , the total annual stock cost is
£ 700,000 / 141 = £ 7.00 * 709 = £ 4963
Quantity per order is 100,000 / n
The maximum stock between orders is also 100,000 / n
So Cost per order = £35 + £7.00 ( (1/n) (100,000)/n )
Since there are n orders, the total cost per year is
C(n) = £35 n + £ 700,000 / n
Differentiating w.r.t. n,
C'(n) = 35 - 700,000 / n^2
Optimum value obtained at n such that C'(n) = 0
Therefore, 700,000 / n^2 = 35
==> n^2 = 700,000 / 35 = 20,000
==> n = sqrt(20,000) = sqrt(2) *100 = 141.4
===> n = 141
Therefore optimum quantity = 100,000/ 141 = 709 litres.
At this value of n , the total annual stock cost is
£ 700,000 / 141 = £ 7.00 * 709 = £ 4963