Maximum Product of Non-Negative Numbers ¹

RN/01/14

W. B. Langdon W.Langdon@cs.ucl.ac.uk
Computer Science, University College, London,
Gower Street, London, WC1E 6BT, UK

14 January 2001

Suppose we wish to maximise

 

$$\prod_{i=1}^{T} P_{i}$$ (1)

subject to the conditions T>0, $P_{i}\ge 0$, $\sum_{i=1}^{T} P_{i} = C$ and $C \ge 0$.

Since logarithm is a monotonically increasing function for the range of P of interest we can maximise (1) by maximising its logarithm. We do this by differentiating w.r.t. P_i for 1<i<T subject to the constraints. The maximum value of (1) occurs when each of the partial derivatives is zero.

$$\log \left(\prod_{i=1}^{T} P_{i} \right) \log \left( C - \sum_{i=2}^{T}P_i \right) + \sum_{i=2}^{T}\log P_i$$ =

$$\stackrel{\textstyle\partial \log \prod_{i=1}^{T} P_{i} } {\textstyle\partial P_i\hfill} -\frac{1}{\left( C - \sum_{i=2}^{T}P_i \right)} + \frac{1}{P_i}$$ =

Setting each partial derivative to zero (and noting $P_i\ne 0$) yields

$$-\frac{1}{\left( C - \sum_{i=2}^{T}P_i \right)} + \frac{1}{P_i}$$ = 0

$$-\left( C - \sum_{i=2}^{T}P_i \right) + P_i$$ = 0

-P₁ + P_i = 0

P_i = P₁

That is the P_i when (1) is maximal are all equal. Since they still have to sum to C every P_i =1/C.