Maximizing A Product With A Fixed Sum

Why Equal Shares Maximize a Product

A fixed total can be divided among several positive quantities in many ways. If the goal is to maximize their product, how should the total be divided? The answer is that equal quantities give the largest product. Differentiation shows why this is true for two quantities, and a pairwise argument extends the conclusion to any finite number.

This result concerns a product, not a sum: when the sum is fixed, every feasible allocation has the same sum, but the products can differ substantially. The result applies when the quantities may be any positive real numbers; it does not assume they must be whole numbers.

Since the sum is fixed, the variables are constrained: choosing \(n-1\) quantities determines the last one. The feasible allocations form the positive part of a simplex. We will first examine the two-quantity case, then prove the general result.

Two Quantities: Differentiate the Product

Suppose two positive quantities \(x\) and \(y\) have fixed sum \(S\). Eliminate \(y\) using \(y=S-x\). Positivity requires \(0<x<S\), and the product becomes a function of one variable:

Its derivative is \(P'(x)=S-2x\). The only critical point in the feasible interval is \(x=S/2\), which also gives \(y=S/2\). The second derivative is \(P''(x)=-2<0\), so the product curves downward there. At either limiting endpoint, one quantity tends to zero and so does the product. Thus the equal split is the global maximum.

The General Result

The same conclusion holds for any finite number of positive quantities. To prove it, consider the product while varying just two quantities and keeping their sum, as well as all the other quantities, fixed. If the two selected quantities are unequal, replacing them with their average increases their product. This gives a direct way to rule out any unequal allocation as a maximum.

Why the Theorem Is True

We give a complete argument, including why a maximum exists and why it must occur at equal quantities. First allow quantities to be zero temporarily and consider the closed feasible set

This set is closed and bounded, hence compact, and the product \(P=x_1\cdots x_n\) is continuous. By the extreme value theorem, it attains a maximum on \(K\). The equal allocation belongs to \(K\) and has positive product \((S/n)^n\). Any point on the boundary of \(K\) has at least one zero quantity and therefore product zero. Consequently, a maximizing allocation is interior: all its quantities are positive. In particular, it is a feasible allocation for the original positive-quantity problem.

Now choose any two quantities in a maximizing allocation, and call them \(a\) and \(b\). Hold all the other quantities fixed. Write \(c=a+b\); the two chosen quantities may be replaced by \(t\) and \(c-t\), where \(0\leq t\leq c\), without changing the total sum. The factor contributed by this pair is

Differentiation gives \(q'(t)=c-2t\). Its only critical point is \(t=c/2\), and \(q''(t)=-2<0\). Equivalently, for every \(t\) in the interval, completing the square gives \(q(t)=c^2/4-(t-c/2)^2\), so the pair product is largest when both quantities equal \(c/2\).

To see strictly why unequal quantities cannot occur at a maximum, compare the original pair product with the product after averaging:

If \(a\ne b\), this difference is positive. The product of the other \(n-2\) quantities is also positive (and, when there are no other quantities, the remaining factor is \(1\)). Therefore replacing \(a,b\) by their average strictly increases the total product while preserving the sum. That contradicts the assumption that the allocation maximizes the product. Since the argument applies to every pair, every pair of quantities at a maximum must be equal. Thus all quantities are \(S/n\), and substitution gives the maximum value \((S/n)^n\). The same reasoning shows uniqueness: any unequal allocation cannot be a maximum.

Applying the Result

A Common Differentiation Error

When differentiating a product written as a function of one variable, forgetting the derivative of one factor gives the wrong critical point. For example, under the constraint \(x+y=12\), write \(P(x)=x(12-x)\). A common incorrect calculation is \(P'(x)=12-x\), as if the second factor were constant. Setting this equal to zero gives \(x=12\), where the product is actually zero.

The product rule, or expansion first, gives the correct derivative:

Thus \(12-2x=0\) gives \(x=6\), and \(y=12-6=6\). The product is \(36\), while the limiting endpoint products at \(x=0\) and \(x=12\) are both zero. The correction is to differentiate the entire objective accurately, then check that the critical point is feasible and compare it with the relevant boundary behavior.

Why the Result Matters

The theorem turns a problem with many possible allocations into a simple calculation: divide the total by the number of quantities, then multiply that common value by itself \(n\) times. It is useful whenever a product is the objective and the only constraint is a fixed sum. If the quantities have units, each factor contributes one unit to the product, so the product has the corresponding units raised to the \(n\)th power.

The assumptions matter. The total must be positive, the number of quantities must be fixed, and the quantities may vary continuously over positive real values. Additional restrictions—such as different lower bounds, different costs for each quantity, or a requirement that quantities be integers—change the feasible set and may change the answer. The equal-allocation result applies to the stated fixed-sum product problem, not automatically to every allocation problem.