Find Two Positive Real Numbers Whose Product Is A Maximum

To find two positive realnumbers whose product is a maximum, you need to consider a simple yet powerful mathematical principle that links their sum to their product. When the sum of the two numbers is fixed, the product reaches its highest possible value when the numbers are equal, a result that can be derived using calculus or the AM‑GM inequality. This article walks you through the reasoning, the steps, and the practical implications of maximizing a product under a constant sum.

Understanding the Problem

The Unconstrained Case Is Trivial

If no restriction is placed on the two numbers, their product can be made arbitrarily large by letting either factor grow without bound. In that scenario, there is no finite maximum; the product can approach infinity. Therefore, to obtain a meaningful answer, a constraint must be introduced. The most common constraint in elementary problems is a fixed sum, denoted as

[x + y = S, ]

where (S) is a given positive constant and (x, y > 0).

Why a Constraint Matters

A constraint transforms an open‑ended optimization problem into a well‑defined one. It forces the search space to a line segment in the first quadrant of the coordinate plane, allowing a genuine maximum to exist. Without such a bound, the notion of “maximum” loses its practical meaning.

The Role of Constraints

Fixed Sum, Variable Product Consider the equation (x + y = S). Solving for (y) gives (y = S - x). Substituting into the product (P = xy) yields a single‑variable function

[ P(x) = x(S - x) = Sx - x^{2}. ]

This quadratic function opens downward (its coefficient of (x^{2}) is negative), guaranteeing a single peak—a maximum—at the vertex of the parabola.

Fixed Product, Variable Sum

Conversely, if the product is fixed and the sum is to be minimized, the same reasoning applies but with the roles of (x) and (y) swapped. The symmetry of the problem makes the analysis interchangeable.

Using Calculus to Maximize the Product

Derivative Test To locate the maximum of (P(x) = Sx - x^{2}), differentiate with respect to (x):

[ P'(x) = S - 2x. ]

Setting the derivative equal to zero gives the critical point

[ S - 2x = 0 \quad \Longrightarrow \quad x = \frac{S}{2}. ]

Because the second derivative (P''(x) = -2) is negative, this critical point is indeed a maximum. Substituting back, [ y = S - x = S - \frac{S}{2} = \frac{S}{2}. ]

Thus, the product is maximized when both numbers are equal, each being half of the prescribed sum.

Verifying the Maximum Value

The maximal product is

[ P_{\text{max}} = \left(\frac{S}{2}\right)!\left(\frac{S}{2}\right) = \frac{S^{2}}{4}. ]

This value is independent of the actual magnitude of (S); it scales with the square of the sum.

Applying the AM‑GM Inequality

Statement of the Inequality

For any two non‑negative real numbers (a) and (b),

[ \frac{a + b}{2} \ge \sqrt{ab}, ]

with equality iff (a = b).

Deriving the Maximum

Re‑arranging the inequality gives

[ ab \le \left(\frac{a + b}{2}\right)^{2}. ]

If (a + b = S), then

[ ab \le \left(\frac{S}{2}\right)^{2} = \frac{S^{2}}{4}. ]

The right‑hand side is exactly the product obtained via calculus. Hence, the AM‑GM inequality provides a quick, non‑calculus proof that the product attains its maximum when the two numbers are equal.

Step‑by‑Step Solution

1. Identify the constraint.
   Determine the fixed sum (S) (or any other condition) that links the two variables.

2. Express one variable in terms of the other.
   From (x + y = S), write (y = S - x).

3. Form the product function.
   Compute (P(x) = x(S - x) = Sx - x^{2}).

4. **Find the

Maximize the Product

5. Differentiate the product function.
   Calculate (P'(x) = S - 2x).

6. Find the critical points.
   Set (P'(x) = 0) and solve for (x). In our example, this yields (x = \frac{S}{2}).

7. Verify that the critical point is a maximum.
   Calculate the second derivative, (P''(x) = -2). Since (P''(x) < 0), the critical point corresponds to a maximum.

8. Determine the optimal values of the variables.
   Substitute (x = \frac{S}{2}) back into the equation (y = S - x) to find the corresponding value of (y). In this case, (y = S - \frac{S}{2} = \frac{S}{2}).

9. Calculate the maximum product.
   Substitute the optimal values of (x) and (y) into the product function (P(x)) to find the maximum product value. This results in (P_{\text{max}} = \left(\frac{S}{2}\right)!\left(\frac{S}{2}\right) = \frac{S^{2}}{4}).

This systematic approach, combining calculus and the AM-GM inequality, provides a robust method for maximizing a product subject to a fixed sum constraint. It’s a powerful illustration of how seemingly abstract mathematical concepts can be applied to solve practical problems. The key takeaway is that symmetry plays a crucial role – when the constraint allows for equal values, the product will invariably be maximized at that point.

Beyond Simple Sums: Generalizations

While the example above focused on a fixed sum, the principles extend to other constraints. Consider a fixed difference, (y - x = D). We can express (y = x + D) and substitute into the product (P = xy) to obtain (P(x) = x(x + D) = x^2 + Dx). Differentiating and setting the derivative to zero yields (P'(x) = 2x + D = 0), so (x = -\frac{D}{2}). The second derivative, (P''(x) = 2), is positive, indicating a minimum. Therefore, to maximize the product with a fixed difference, the two numbers must be equal, as in the sum case.

Similarly, constraints involving ratios or other relationships can be transformed into a similar form, allowing the same calculus and AM-GM techniques to be applied. The underlying principle remains consistent: symmetry and the properties of the constraint dictate the optimal solution.

Conclusion

The pursuit of maximum values under constraints is a fundamental problem in optimization. By employing tools like calculus and inequalities, we can systematically determine the conditions that yield the highest possible outcome. The AM-GM inequality, in particular, offers a concise and elegant proof of the optimal solution, highlighting the inherent balance between variables when a constraint is imposed. Ultimately, understanding these techniques provides a valuable framework for tackling a wide range of real-world scenarios, from resource allocation and financial modeling to engineering design and even strategic decision-making.