Maximum Product of Non-Negative Numbers 1


W. B. Langdon
Computer Science, University College, London,
Gower Street, London, WC1E 6BT, UK

14 January 2001

Suppose we wish to maximise

$\displaystyle \prod_{i=1}^{T} P_{i}$     (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. Pi for 1<i<T subject to the constraints. The maximum value of (1) occurs when each of the partial derivatives is zero.

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

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

$\displaystyle -\frac{1}{\left( C - \sum_{i=2}^{T}P_i \right)}
\frac{1}{P_i}$ = 0  
$\displaystyle -\left( C - \sum_{i=2}^{T}P_i \right)
P_i$ = 0  
-P1 + Pi = 0  
Pi = P1  

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

W. B. Langdon
2001-01-14 (2 April 2001)